What is Additional Mathematics | The Exponential Equation That Hides a Quadratic

Spot the common exponential base before taking logarithms

The equation 4^x−5·2^x+4=0 contains two exponential expressions that are related: 4^x=(2^x)². That identity turns the original equation into a quadratic in the single positive quantity 2^x. Taking logarithms of the whole sum does not separate its terms, so the correct first question is whether a common base can be found. Write 4 as 2² and inspect the exponents. The highest power is twice the lower one, exactly the structure required for a quadratic substitution. This recognition is the main skill that the example is designed to teach.

Why u=2^x must be positive

For every real x, the exponential 2^x is strictly positive. It may approach zero as x decreases without bound, but it never equals zero or becomes negative. Thus u=2^x comes with the condition u>0. When a substituted quadratic has a negative root, that root is a valid algebraic solution for the polynomial in an unrestricted variable but is impossible for the chosen exponential substitution. For example, 4^x+2^x−2=0 becomes u²+u−2=0, giving u=1 or −2. Only u=1 is permitted, so the original equation has x=0 alone.

The worked quadratic in full detail

Let u=2^x>0. Then 4^x=(2^x)²=u², so 4^x−5·2^x+4=0 becomes u²−5u+4=0. Factor as (u−1)(u−4)=0, giving u=1 or u=4. Restore the original expression: 2^x=1 gives x=0, and 2^x=4 gives x=2. Verify directly: at x=0, 1−5+4=0; at x=2, 16−20+4=0. Both are valid, and there are no other solutions because the quadratic has exactly those two roots and 2^x is one-to-one.

Why the quadratic factorisation is not the final answer

Writing u=1 or u=4 does not answer a question that asks for x. The auxiliary variable is only a temporary representation of the exponential expression. A complete solution returns to 2^x=1 and 2^x=4 and then gives x=0 or x=2. The return step may require logarithms if the positive auxiliary root is not a simple power of the base. For instance, if 2^x=3, the exact answer is x=ln3/ln2. It is not necessary to force a decimal approximation unless the question asks for one.

A quadratic with one admissible root

Solve 4^x+3·2^x−4=0. Put u=2^x>0, giving u²+3u−4=0. Factor as (u+4)(u−1)=0. The quadratic roots are u=−4 and u=1. Reject the negative root because no real power of two is negative. The remaining equation 2^x=1 gives x=0. Substitution into the original yields 1+3−4=0. This example is useful because it separates solving the auxiliary polynomial from deciding which auxiliary roots belong to the original exponential function’s range.

A quadratic with two positive roots and logarithms

Solve 4^x−7·2^x+10=0. Substitute u=2^x>0 to obtain u²−7u+10=0, or (u−2)(u−5)=0. Thus u=2 or u=5. The first gives x=1. The second gives x=log₂5=ln5/ln2. Both are exact answers. The second cannot be simplified to an integer because five is not an integer power of two. A decimal approximation is about 2.322, but the logarithmic expression is the exact solution and is generally preferable when exact values are requested.

A repeated root gives one distinct solution

For 4^x−4·2^x+4=0, let u=2^x>0. Then u²−4u+4=(u−2)²=0, so u=2 and x=1. Although the polynomial has a repeated root, there is only one distinct x-value. On a graph, the original function is (2^x−2)², which is nonnegative and touches the x-axis at x=1 without crossing it. This connects algebraic multiplicity with graphical behaviour. The answer should list x=1 once, not claim that the repeated auxiliary root creates two different real solutions.

When there is no solution for an obvious reason

Consider 4^x+2^x+1=0. Every term on the left is strictly positive for every real x, so their sum cannot be zero. No substitution is necessary. The quadratic method would give u²+u+1=0, whose discriminant is negative, confirming the same conclusion. The positivity check is quicker and also explains the answer. Before beginning a long algebraic procedure, inspect whether the signs of the terms already make the proposed equality impossible.

When there is no solution because both auxiliary roots are negative

Solve 4^x+3·2^x+2=0. Let u=2^x>0. The quadratic is u²+3u+2=(u+1)(u+2)=0, so its roots are u=−1 and u=−2. Both violate the positive range of 2^x. Therefore the exponential equation has no real solutions. This example differs from the negative-discriminant case: the polynomial has two real roots, but neither is permitted by the substitution. Students should distinguish these two reasons for an empty solution set.

A base-three version

Consider 9^x−10·3^x+9=0. Since 9^x=(3^x)², put u=3^x>0. Then u²−10u+9=(u−1)(u−9)=0, so u=1 or 9. Return to the original base: 3^x=1 gives x=0, while 3^x=9 gives x=2. Both answers satisfy the original equation. The method is not limited to base two. What matters is that the exponential terms can be rewritten as powers of one positive base other than one.

An equation with a fractional power of the base

Solve 2^x+2^(−x)=5/2. Let u=2^x>0, so 2^(−x)=1/u. The equation becomes u+1/u=5/2. Multiply by 2u, which is safe because u is positive and nonzero, to obtain 2u²−5u+2=0. Factor as (2u−1)(u−2)=0. Thus u=1/2 or u=2, giving x=−1 or x=1. Direct substitution confirms that 2+1/2=5/2 for x=1 and 1/2+2=5/2 for x=−1. Reciprocal exponentials can therefore hide a quadratic too.

A symmetry check for reciprocal exponentials

The function F(x)=2^x+2^(−x) is even because F(−x)=2^(−x)+2^x=F(x). Therefore its real solutions at a level greater than the minimum usually occur in opposite pairs. In the preceding example, x=−1 and x=1 are consistent with that symmetry. Moreover, if u=2^x>0, then u+1/u≥2, since (u−1)²≥0 implies u²+1≥2u. Thus the level 5/2 is attainable, whereas an equation F(x)=3/2 has no real solution. This inequality provides an independent conceptual check.

A substitution using an exponential shift

Solve 4^(x+1)−9·2^(x+1)+8=0. Put u=2^(x+1)>0. Because 4^(x+1)=[2^(x+1)]²=u², the equation becomes u²−9u+8=(u−1)(u−8)=0. Thus 2^(x+1)=1 or 8. The first gives x+1=0 and x=−1; the second gives x+1=3 and x=2. Choosing the whole repeated exponential 2^(x+1) as u avoids unnecessarily expanding exponents. The substitution should follow the repeated structure in the equation.

A base-change identity for exact solutions

Suppose a valid auxiliary root gives 3^x=7. Taking natural logarithms of both sides is permitted because both sides are positive. Using ln(3^x)=x ln3, we obtain x=ln7/ln3. The same number can be written log₃7. This logarithmic step is different from taking the logarithm of a sum of unrelated terms. Here the entire positive exponential has been isolated first, so the logarithm law applies correctly. The exact expression is often clearer than a prematurely rounded decimal.

Why logarithms cannot be distributed across a sum

It is false that ln(a+b)=ln a+ln b. For example, with a=b=1, the left side is ln2 while the right side is ln1+ln1=0. Therefore taking the logarithm of 4^x−5·2^x+4 and trying to split it term by term is invalid. The negative middle term also creates potential domain problems if it is logged separately. The valid laws are ln(ab)=ln a+ln b and ln(a/b)=ln a−ln b for positive a and b. Recognising the difference prevents a common algebraic error.

Use the quadratic discriminant before factorising

For an equation a u²+b u+c=0, the discriminant Δ=b²−4ac tells us whether the auxiliary quadratic has two distinct real roots, one repeated real root or no real roots. However, it does not by itself tell us how many exponential solutions exist. Each real auxiliary root must also satisfy u>0. A quadratic with two real negative roots gives no exponential solutions; a quadratic with one positive and one negative root gives one. Thus the discriminant is a preliminary algebraic test, while the positivity restriction completes the original problem.

A quadratic that is easiest by the formula

Solve 4^x−3·2^x+1=0. Let u=2^x>0. The quadratic u²−3u+1=0 has roots u=(3±√5)/2. Both roots are positive because 3>√5. Therefore x=ln[(3+√5)/2]/ln2 or x=ln[(3−√5)/2]/ln2. These are exact solutions. The two auxiliary roots multiply to one, so their logarithms are negatives of each other. This symmetry can be used as an independent check, but it does not replace the derivation.

A quadratic with a zero auxiliary root

Consider 4^x−2^x=0. Substitute u=2^x>0, obtaining u²−u=u(u−1)=0. The polynomial roots are u=0 and u=1. Reject u=0 because 2^x never equals zero for a finite real x. The only solution is 2^x=1, so x=0. This example highlights a subtlety: an exponential can approach zero without ever attaining it. A limit value is not necessarily an admissible solution.

Graph the transformed quadratic conceptually

Write F(x)=4^x−5·2^x+4 as Q(u)=u²−5u+4 with u=2^x>0. The quadratic Q has zeros at u=1 and u=4 and a vertex at u=5/2. As x increases, u increases continuously from values near zero to arbitrarily large values. Consequently F(x) is positive when 0<u<1, negative when 1<u<4 and positive when u>4. Translating these intervals back gives F(x)>0 for x<0 and x>2, and F(x)<0 for 0<x<2. The sign pattern confirms that the graph crosses the x-axis at exactly x=0 and x=2.

Locate the minimum without differentiating

The quadratic Q(u)=u²−5u+4 can be completed to a square: Q(u)=(u−5/2)²−9/4. Its minimum is −9/4 at u=5/2. Because u=2^x covers all positive real numbers, that minimum is attained at x=log₂(5/2). Thus the original exponential expression has minimum −9/4. This extension shows how a quadratic substitution can answer more than an equation-solving question: it can also describe the range of an exponential expression. The positivity restriction does not exclude the vertex because 5/2>0.

When the quadratic vertex is outside the permitted range

Consider F(x)=4^x+2^x+1. With u=2^x>0, Q(u)=u²+u+1 has its unconstrained vertex at u=−1/2, outside the permitted positive domain. For u>0, Q is strictly increasing, and its values approach one as u approaches zero from above. Thus F(x)>1 for every real x, with no attained minimum equal to one. A student who reports the quadratic’s unrestricted vertex value as the minimum of the exponential has ignored the substitution domain. The range restriction matters in optimisation as well as equation solving.

A worked exponential inequality

Solve 4^x−5·2^x+4≤0. Put u=2^x>0 and factor (u−1)(u−4)≤0. The quadratic is nonpositive between its roots, so 1≤u≤4. Because 2^x is strictly increasing, this is equivalent to 0≤x≤2. The endpoints are included because the inequality allows equality. For the strict version F(x)<0, the answer would be 0<x<2. This is a natural extension of the equation and reinforces how algebraic sign intervals translate through a monotonic exponential.

An inequality with one negative auxiliary boundary

Solve 4^x+3·2^x−4>0. Let u=2^x>0. Factor u²+3u−4=(u+4)(u−1)>0. Since u+4 is always positive for u>0, the inequality reduces to u−1>0, or u>1. Hence x>0. The other quadratic sign interval u<−4 is not permitted. A student who carries that negative interval back into an exponential inequality has forgotten the positive range of the substitution.

How the base affects the final inequality

If the common base a is greater than one, the function a^x is increasing. Thus a^x>k for k>0 means x>log_a k. If 0<a<1, the exponential is decreasing, and the inequality direction reverses: a^x>k means x<log_a k. This distinction is essential when solving exponential inequalities by substitution. The auxiliary polynomial’s sign analysis occurs in u, but translating a permitted interval in u back to x depends on whether the original exponential increases or decreases.

A decreasing-base worked example

Solve (1/4)^x−(5/2)(1/2)^x+1=0. Let u=(1/2)^x>0. Then (1/4)^x=u², giving u²−(5/2)u+1=0. Multiply by two: 2u²−5u+2=(2u−1)(u−2)=0. Thus u=1/2 or u=2. Since (1/2)^1=1/2 and (1/2)^(−1)=2, the solutions are x=1 and x=−1. The base is below one, but the equation-solving substitution still works; monotonicity matters especially when converting inequalities.

A mixed exponent that is not quadratic

Consider 8^x−7·2^x+6=0. With u=2^x>0, 8^x=(2^x)³=u³, so the equation becomes u³−7u+6=0. This is cubic, not quadratic. It factors as (u−1)(u−2)(u+3)=0. The positive roots are u=1 and u=2, giving x=0 and x=1; u=−3 is invalid. The common-base substitution remains useful even when the transformed polynomial has a higher degree. Students should identify the degree rather than force every exponential expression into a quadratic pattern.

A substitution that fails to simplify

The equation 2^x+3^x=5 does not have both exponential terms as integer powers of one obvious common base. Setting u=2^x leaves 3^x, which is not a simple quadratic expression in u. One can sometimes analyse such equations through monotonicity or numerical methods, but the quadratic substitution taught here does not directly apply. At x=1, 2+3=5, so x=1 is a solution. Since 2^x+3^x is strictly increasing for real x, this solution is unique. Recognising when a method does not fit is part of mathematical mastery.

A practical substitution decision tree

First inspect whether all exponential terms use powers of one positive base a≠1. Next look for exponents in a ratio such as 2x and x, or x and −x. Define u=a^x or another repeated positive exponential expression. Rewrite every term in u, then solve the resulting polynomial or rational equation. Reject u≤0, because the original exponential is strictly positive. Back-substitute each remaining root and use logarithms only after isolating a positive exponential. Finally check each candidate in the original equation and state exact values where possible.

Diagnose an incorrect logarithm step

Suppose a student writes ln(4^x−5·2^x+4)=x ln4−5x ln2+ln4. The first error is the unsupported splitting of the logarithm of a sum and difference. The correct path is to recognise 4^x=(2^x)² and solve the auxiliary quadratic. If a student instead writes u=2^x and keeps u=−2 from a quadratic, the first error is forgetting the positive range of the substitution. If the student stops at u=4, the error is failure to return to x. These mistakes need separate explanations rather than a generic instruction to practise more.

A complete exam-style worked solution

Question: Solve 9^x−12·3^x+27=0. Since 9^x=(3^x)², let u=3^x>0. The equation becomes u²−12u+27=0, which factors as (u−3)(u−9)=0. Thus u=3 or u=9, giving 3^x=3 or 3^x=9. Therefore x=1 or x=2. Check x=1: 9−36+27=0. Check x=2: 81−108+27=0. Both are valid. The solution is complete because all auxiliary roots were considered, both satisfy positivity, and both were translated back into the original variable.

A second worked solution with an exact logarithm

Question: Solve 4^x−6·2^x+5=0. Let u=2^x>0, giving u²−6u+5=(u−1)(u−5)=0. Therefore u=1 or u=5. Back-substitution gives x=0 or x=ln5/ln2. Both roots are positive in u, so neither is rejected. Check x=0 directly: 1−6+5=0. For the second, 2^x=5 and 4^x=25, so the original left side is 25−30+5=0. This exact verification avoids approximating the logarithm before checking.

Independent practice with answers

Question A: Solve 4^x−3·2^x+2=0. Answer x=0 or x=1. Question B: Solve 9^x−4·3^x+3=0. Answer x=0 or x=1. Question C: Solve 4^x+3·2^x+2=0. Answer: no real solutions because both auxiliary roots are negative. Question D: Solve 2^x+2^(−x)=2. Answer x=0, since the auxiliary equation is (u−1)²=0. Question E: Solve 4^x−5·2^x+4<0. Answer 0<x<2. Each question should show the substitution and its positivity condition.

A more demanding practice set

Question F: Solve 4^x−8·2^x+12=0. Let u=2^x>0. Then (u−2)(u−6)=0, so x=1 or x=ln6/ln2. Question G: Solve 9^x+3^x−2=0. The auxiliary roots are one and negative two, so x=0 only. Question H: Solve 4^x−2·2^x+1=0. The auxiliary equation is (u−1)²=0, giving x=0. Question I: Solve 4^x−5·2^x+4≥0. Answer x≤0 or x≥2. These exercises test exact logarithms, rejected roots, repeated roots and inequalities.

How to revise without memorising one equation

Start with a simple factorable example using base two. Next use base three. Then add a negative auxiliary root, a repeated root and a zero auxiliary root. Follow with a positive root that requires an exact logarithm. After that, introduce reciprocal exponentials and inequalities. The learner should explain why u>0 at every stage. Only then mix in examples where the common-base substitution is cubic or unsuitable. This progression makes the choice of method deliberate instead of automatic.

What parents can ask after the lesson

Ask the learner to identify the repeated exponential expression before writing any algebra. Why is 4^x the square of 2^x? Why must the new variable be positive? What would a negative quadratic root mean? Why can the logarithm of a positive isolated exponential be taken, while the logarithm of a sum cannot be split? Can the learner check the final x-values in the original equation? These questions expose the reasoning behind the method and help distinguish genuine understanding from copying a factorisation.

Frequently asked questions on exponential substitution

Must the common base be two? No, any suitable positive base other than one can work. Why is the auxiliary variable positive? Real exponentials with positive bases are strictly positive. What if the quadratic root is zero? Reject it. What if the root is five? Use an exact logarithm if five is not a convenient power of the base. Does every exponential equation become quadratic? No; some become cubic or do not simplify under one common-base substitution. Can the same method solve inequalities? Yes, but translate intervals using the exponential’s monotonicity.

Final accuracy checklist and syllabus boundary

Confirm that the exponential terms have been rewritten using valid index laws. State the positive range of the substituted variable. Solve the transformed equation correctly and reject every nonpositive auxiliary root. Back-substitute into the original exponential, using exact logarithms where necessary. Check each final answer in the original equation and do not split logarithms across addition. For inequalities, track strict endpoints and whether the exponential base is increasing or decreasing. Consult the official SEAB 2027 G3 Additional Mathematics K341 syllabus for the exact examinable scope; some cubic and optimisation examples here are extensions rather than a claim about examination requirements.

An equation with a squared shifted exponential

Solve (2^x+1)²−5(2^x+1)+6=0. Put u=2^x+1. Because 2^x>0, we have u>1, not merely u>0. The quadratic becomes u²−5u+6=(u−2)(u−3)=0. Both roots satisfy u>1. If u=2, then 2^x=1 and x=0. If u=3, then 2^x=2 and x=1. Check in the original: at x=0 the bracket is two, giving 4−10+6=0; at x=1 the bracket is three, giving 9−15+6=0. The permitted range of a shifted exponential must be written accurately.

A shifted substitution with a rejected positive root

Now solve (2^x+1)²−(5/2)(2^x+1)+1=0. Set u=2^x+1>1. Multiply the auxiliary equation by two: 2u²−5u+2=(2u−1)(u−2)=0. The roots are u=1/2 and u=2. Although 1/2 is positive, it violates the stronger restriction u>1 and must be rejected. The remaining u=2 gives 2^x=1 and x=0. This example demonstrates why it is not enough to memorise that every auxiliary variable is positive: the exact range depends on the chosen expression.

How to use a solution count as a check

For a quadratic in u=a^x with a>0 and a≠1, there can be at most two distinct real x-solutions because the quadratic has at most two distinct real roots and a^x is one-to-one. This upper bound applies to that specific substitution structure. It does not apply to every exponential equation: a cubic in a^x can have three positive roots and therefore three distinct real solutions. Nor does it mean that two quadratic roots always give two x-values; zero or negative auxiliary roots are inadmissible. Counting roots is a useful reasonableness check only after the equation’s transformed degree and domain are understood.

A graphical check using monotonicity

For u=2^x, the mapping from x to u is continuous and strictly increasing. Every positive u corresponds to exactly one x, so solving the transformed quadratic for positive u-values loses no real solutions. When the quadratic factors into (u−1)(u−4), the sign changes at u=1 and u=4 correspond exactly to x=0 and x=2. This is why the graph of the original exponential expression crosses the axis at those values. The transformation is not a trick: it re-labels a positive, continuously increasing input and preserves the order of its values.

A final mini-test before submission

Given 4^x−6·2^x+8=0, write u=2^x>0. The transformed quadratic is u²−6u+8=(u−2)(u−4), so u=2 or u=4. Return to x to obtain x=1 or x=2. Check x=1: 4−12+8=0. Check x=2: 16−24+8=0. Now change the final constant from eight to nine: the quadratic becomes (u−3)²=0, giving the single solution x=ln3/ln2. Explain why the answer count changes, why the logarithm is exact and why no negative auxiliary root is permitted.