P2 5.3 - Solving exponential equations | Edexcel A-Level International Mathematics

Solve equations in which the unknown occurs in an exponent, using a common base or logarithms. You will derive change of base, calculate approximations without losing accuracy, and keep exact answers when finding graph intersections.

Same-base equations

Where do $y=2^{2x+1}$ and $y=10\times2^x$ meet? Trying integer inputs will miss the intersection. We will develop a method that gives both coordinates exactly.

Start with the simpler fact $2^t=2^5$ . Because $2^x$ is strictly increasing, the two outputs can be equal only when $t=5$ . A decay exponential is strictly decreasing, but also takes each output only once, so the same reasoning applies.

You already know

$(a^m)^n=a^{mn}$ , $a^m a^n=a^{m+n}$ , and $a^{-m}=1/a^m$ for a positive base. These laws let you rewrite different-looking powers with one common base.

For $a>0$ , $a\ne1$ ,

a^{f(x)}=a^{g(x)}\quad\Longrightarrow\quad f(x)=g(x).

First make the bases genuinely equal. You cannot simply equate the exponents in $4^x=8^2$ .

Worked example 1: rewriting both sides

Question. Solve $4^{x+1}=8$ .

Plan. Write both sides as powers of $2$ , then equate exponents.

\begin{aligned} (2^2)^{x+1}&=2^3\\ 2(x+1)&=3 &&\text{equal bases}\\ x&=\frac12. \end{aligned}

Answer. $x=1/2$ .

Check. $4^{1/2+1}=4^{3/2}=(\sqrt4)^3=8$ .

Change only the right-hand side to its reciprocal. The corresponding exponent changes sign.

Worked example 2: a reciprocal target

Question. Solve $4^{x+1}=1/8$ .

Plan. Use base $2$ again, writing the reciprocal as a negative power.

\begin{aligned} 2^{2(x+1)}&=2^{-3}\\ 2x+2&=-3\\ x&=-\frac52. \end{aligned}

Answer. $x=-5/2$ .

Check. The exponent on the original base is $-5/2+1=-3/2$ , so $4^{-3/2}=1/(\sqrt4)^3=1/8$ .

Brackets prevent a common index slip.

Watch out

$4^{x+1}=2^{2(x+1)}=2^{2x+2}$ , not $2^{2x+1}$ . The outer power multiplies the entire exponent.

Choose a common base for this reciprocal target.

Using logarithms

When the target is not a convenient power of the base, a logarithm brings the unknown exponent down where ordinary algebra can reach it.

For $5^t=12$ , take base- $10$ logarithms on both sides:

\log_{10}(5^t)=\log_{10}12 \quad\Longrightarrow\quad t\log_{10}5=\log_{10}12.

The power law is the key step. Dividing by $\log_{10}5\ne0$ isolates $t$ . Any single valid logarithm base works, provided you use it consistently.

For an equation $a^t=b$ with $a>0$ , $a\ne1$ , a real solution requires $b>0$ . If $b\le0$ , positivity of the exponential already proves there is no real solution; do not take a logarithm of that target.

Worked example 3: an exponent with no common-base shortcut

Question. Solve $5^x=12$ , giving an exact expression and a value to three decimal places.

Plan. Take logarithms, use the power law and divide by the logarithm of the base.

\begin{aligned} \log_{10}(5^x)&=\log_{10}12\\ x\log_{10}5&=\log_{10}12\\ x&=\frac{\log_{10}12}{\log_{10}5}. \end{aligned}

Answer. $x=\log_{10}12/\log_{10}5\approx1.544$ .

Check. Since $5<12<25$ , the exponent must lie between $1$ and $2$ . Substituting the unrounded quotient into $5^x$ gives $12$ ; using $1.544$ gives approximately $12.001$ .

Change only the exponent from $x$ to $2x-1$ . Logarithms isolate the entire bracket first.

Worked example 4: a linear expression in the exponent

Question. Solve $5^{2x-1}=12$ , giving an exact expression and a value to three decimal places.

Plan. Apply the power law to $2x-1$ , then solve the resulting linear equation.

\begin{aligned} (2x-1)\log_{10}5&=\log_{10}12\\ 2x-1&=\frac{\log_{10}12}{\log_{10}5}\\ x&=\frac12\left(1+\frac{\log_{10}12}{\log_{10}5}\right). \end{aligned}

Answer. $x=\frac12(1+\log_{10}12/\log_{10}5)\approx1.272$ .

Check. The unrounded value makes $2x-1$ equal to the exponent in Example 3. Thus $5^{2x-1}=12$ , independently of any final rounding.

Keep full calculator precision until the requested final answer.

Watch out

$\log_{10}(5^{2x-1})=(2x-1)\log_{10}5$ . Writing $2x-\log_{10}5$ loses the bracket and changes the equation. Also, $\log_{10}(6\times3^x)=\log_{10}6+x\log_{10}3$ , not a product of logarithms.

Use the same sequence of steps with a new linear exponent.

Change of base

A calculator need not have a key for your chosen base: every logarithm can be expressed using another valid base.

Let $t=\log_a b$ . Its meaning is $a^t=b$ . Taking base- $c$ logarithms gives $t\log_c a=\log_c b$ , so division gives the change-of-base formula.

Change of base

For $a,c>0$ , $a,c\ne1$ and $b>0$ ,

\log_a b=\frac{\log_c b}{\log_c a}.

The argument's logarithm goes on top; the original base's logarithm goes below.

In the formula booklet.

In particular, the exact answer in Example 3 is $\log_5 12$ . Its base- $10$ quotient is just another exact notation for the same number. A base between zero and one is valid; its base- $10$ logarithm is negative.

Worked example 5: calculating a logarithm in a new base

Question. Find $\log_4 7$ to four decimal places.

Plan. Change to base $10$ , retaining all digits until the end.

\log_4 7=\frac{\log_{10}7}{\log_{10}4} =1.403677\ldots

Answer. $1.4037$ .

Check. $4^1<7<4^2$ , so the answer lies between $1$ and $2$ . Raising $4$ to the unrounded quotient returns $7$ .

Change only the original base to its reciprocal. The numerator stays the same and the denominator changes sign.

Worked example 6: a decay base

Question. Find $\log_{1/4}7$ to four decimal places.

Plan. Use the same formula with $\log_{10}(1/4)=-\log_{10}4$ .

\log_{1/4}7 =\frac{\log_{10}7}{\log_{10}(1/4)} =-\frac{\log_{10}7}{\log_{10}4} =-1.403677\ldots

Answer. $-1.4037$ .

Check. If $t=\log_4 7$ , then $(1/4)^{-t}=4^t=7$ . The new exponent must therefore be the negative of Example 5's value.

The order in the quotient matters.

Watch out

$\log_4 7$ is $\log 7/\log4$ , not $\log4/\log7$ . A quick bound from powers of the original base catches an upside-down quotient.

Keep the negative denominator when the base is below one.

Exact and approximate answers

An exact logarithm preserves information that a rounded decimal loses, especially when you use the answer in another exponential expression.

For example, $2^{\log_2 5}=5$ exactly. There is no need to calculate $\log_2 5$ before using it. An exact answer is useful when a question asks for a logarithmic coordinate or an integer expression for another coordinate; approximate only when requested.

Problem-solving

At an intersection, equate the two expressions for $y$ . Simplify powers before taking logarithms: division by $a^x$ is valid because it is positive, and may leave a much simpler equation.

This supplies the method for the opening problem.

Worked example 7: exact coordinates of an intersection

Question. Find the exact intersection of $y=2^{2x+1}$ and $y=10\times2^x$ , writing the first coordinate as a single logarithm.

Plan. Equate the outputs, divide by the positive factor $2^x$ , then reuse the resulting exact power to find $y$ .

\begin{aligned} 2^{2x+1}&=10\times2^x\\ 2^{x+1}&=10 &&\text{divide by }2^x\\ 2^x&=5 &&\text{divide by }2\\ x&=\log_2 5\\ y&=10\times5=50. \end{aligned}

Answer. $(\log_2 5,50)$ .

Check. Using the other curve gives $2^{2x+1}=2(2^x)^2=2\times25=50$ . Both curves have the same height, and $2^x=5$ has only one real solution.

Change only the coefficient of the second curve from $10$ to $14$ . Its exact height changes as well as the logarithm's argument.

Worked example 8: a different coefficient

Question. Find the exact intersection of $y=2^{2x+1}$ and $y=14\times2^x$ , writing the first coordinate as a single logarithm.

Plan. Simplify the common exponential factor before evaluating either coordinate.

\begin{aligned} 2^{2x+1}&=14\times2^x\\ 2^{x+1}&=14\\ 2^x&=7\\ x&=\log_2 7,\qquad y=14\times7=98. \end{aligned}

Answer. $(\log_2 7,98)$ .

Check. The first curve gives $2(2^x)^2=2\times49=98$ . Since $4<7<8$ , the first coordinate lies between $2$ and $3$ , consistent with the two heights.

These calculations answer the opening puzzle without a search through decimal inputs. If an approximation were requested, evaluate the exact logarithm only at the final step.

Exam tip

An exact-answer instruction calls for an exact form such as $\log_2 7$ , rather than only a decimal. When all stages are requested, show the index or logarithm steps that lead to it.

Change the coefficient once more and retain the exact power for the height.

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P2 5.3 - Solving exponential equations

Same-base equations

Using logarithms

Change of base

Exact and approximate answers

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