ME6 - Solving exponential and logarithmic equations | OCR A-Level Mathematics B (MEI)

Solve exponential equations by choosing logarithms, equal bases or a positive substitution. Check logarithm domains, interpret numerical intersections, and solve related inequalities using the direction of the exponential or logarithm.

Taking logarithms to find an exponent

Can changing only an exponential’s base reverse a whole solution interval? At $x=3$ , $2^x=8$ but $(1/2)^x=1/8$ . One rises as the input rises; the other falls. First find equality accurately, then use direction to decide the interval.

You already know

For $u>0$ , $\ln(u^r)=r\ln u$ . Natural logarithms undo natural exponentials. Positive-base exponentials other than base 1 are one-to-one; reciprocal bases reverse the direction of change.

An exponent becomes a multiplier when you take logarithms of two positive sides.

Recovering an exponent

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

a^x=b\quad\Rightarrow\quad x\ln a=\ln b\quad\Rightarrow\quad x=\frac{\ln b}{\ln a}.

This is a consequence of the power law, obtained by taking logarithms of both sides.

Not in the formula sheet: learn it.

You may use a consistent base-ten logarithm instead. Keep exact logarithms until the end, rather than rounding each one separately.

Worked example 1: a linear exponent

Question. Determine the exact solution of $3^{x+2}=10$ .

Plan. Take natural logarithms and solve for the complete exponent.

\begin{aligned} (x+2)\ln3&=\ln10 &&\text{power law}\\ x+2&=\frac{\ln10}{\ln3} &&\text{divide by nonzero }\ln3\\ x&=\frac{\ln10}{\ln3}-2 &&\text{isolate }x \end{aligned}

Why this step? The brackets keep both terms of $x+2$ multiplying the logarithm.

Answer. $x=\ln10/\ln3-2$ .

Check. The resulting exponent is $\ln10/\ln3$ , so the logarithm of the left side is exactly $\ln10$ .

Change only the right side from the constant 10 to $10^x$ .

Worked example 2: different bases on both sides

Question. Determine the exact solution of $3^{x+2}=10^x$ .

Plan. Take logarithms of both sides, expand, and collect the terms containing $x$ .

\begin{aligned} (x+2)\ln3&=x\ln10 &&\text{power law on both sides}\\ 2\ln3&=x(\ln10-\ln3) &&\text{collect terms}\\ x&=\frac{2\ln3}{\ln(10/3)} &&\text{solve} \end{aligned}

Why this step? Different bases are manageable once each variable exponent becomes a multiplier.

Answer. $x=2\ln3/\ln(10/3)$ .

Check. Numerically $x\approx1.82498$ ; using the exact form makes $(x+2)\ln3-x\ln10=0$ .

For a natural exponential, isolate it first. Thus $4\mathrm e^{2x-1}=7$ gives $2x-1=\ln(7/4)$ , so $x=(1+\ln(7/4))/2$ . Likewise $2^{x+1}=3^{x-1}$ leads to $x=\ln6/\ln(3/2)$ .

Watch out

The equation $a^x=b$ with $a>0$ cannot have a real solution if $b\le0$ . Base 1 is exceptional: $1^x=1$ holds for all real inputs, while $1^x=b\ne1$ has no solution. Dividing by $\ln1$ is invalid.

Try a case where the exact exponent turns out to be negative.

Equal bases and exponential substitutions

A common base may remove the need for logarithms, or reveal a familiar quadratic hidden inside an exponential equation.

For $3^{2x-1}=27=3^3$ , equal outputs imply $2x-1=3$ , so $x=2$ . The argument works for a decreasing positive-base exponential too, because it is still one-to-one. Base 1 is excluded. For $2^{2x}=(2^x)^2$ , the substitution $u=2^x>0$ gives a quadratic in $u$ ; similarly, $2^{x+1}=2u$ . These index identities are not supplied on the formula sheet.

Worked example 3: a quadratic in a positive quantity

Question. Determine all real solutions of $2^{2x}-5(2^x)-6=0$ .

Plan. Let $u=2^x>0$ , factorise, and convert positive candidates back to $x$ .

\begin{aligned} u^2-5u-6&=(u-6)(u+1)=0 &&\text{substitute and factorise}\\ u&=6\text{ or }-1 &&\text{quadratic roots}\\ 2^x&=6 &&\text{reject negative }u\\ x&=\frac{\ln6}{\ln2} &&\text{take logarithms} \end{aligned}

Why this step? The quadratic can have negative roots, but the substituted quantity cannot be negative.

Answer. $x=\ln6/\ln2$ only.

Check. The starting expression becomes $36-30-6=0$ .

Change only the constant term, from $-6$ to 6.

Worked example 4: keeping both positive roots

Question. Determine all real solutions of $2^{2x}-5(2^x)+6=0$ .

Plan. Use the same substitution and retain every positive root.

\begin{aligned} u^2-5u+6&=(u-2)(u-3)=0 &&\text{factorise}\\ u&=2\text{ or }3 &&\text{both positive}\\ x&=1\text{ or }\frac{\ln3}{\ln2} &&\text{solve each exponential} \end{aligned}

Why this step? Returning from each valid $u$ -value is essential; the question asks for $x$ , not $u$ .

Answer. $x=1$ or $x=\ln3/\ln2$ .

Check. The starting left sides are $4-10+6=0$ and $9-15+6=0$ .

With base $\mathrm e$ , the same method solves $\mathrm e^{2x}-3\mathrm e^x-4=0$ : its exponential candidates are $4,-1$ , so only $x=\ln4$ is valid.

Exam tip

For detailed reasoning, explain a rejection by writing that the exponential is strictly positive for all real inputs. A bare “not possible” does not identify the reason.

Change the constant again while keeping the quadratic's other coefficients fixed.

Logarithmic equations and original domains

An algebraic candidate is a logarithmic solution only when every starting argument is positive.

For positive $U,V$ , $\ln U=\ln V$ implies $U=V$ , because $\ln$ is one-to-one. For a numerical right side instead, exponentiate: $\ln(2x+1)=3$ gives $x=(\mathrm e^3-1)/2$ . These inverse properties are not supplied on the formula sheet.

Worked example 5: filtering a quadratic

Question. Solve $\ln(x-2)+\ln(x+1)=\ln10$ .

Plan. Record the full domain, combine logarithms, then check both candidates.

\begin{aligned} x&>2 &&\text{both arguments positive}\\ (x-2)(x+1)&=10 &&\text{product law and injectivity}\\ x^2-x-12&=(x-4)(x+3)=0 &&\text{rearrange and factorise}\\ x&=4\text{ or }-3 &&\text{algebraic candidates} \end{aligned}

Why this step? A positive product can hide two negative starting arguments.

Answer. Only $x=4$ satisfies $x>2$ .

Check. $\ln2+\ln5=\ln10$ . At $-3$ , both starting arguments are negative, so that candidate is invalid.

Change only addition to subtraction between the logarithms.

Worked example 6: an equation with no real solution

Question. Solve $\ln(x-2)-\ln(x+1)=\ln10$ .

Plan. Keep the domain $x>2$ , and use the quotient law.

\begin{aligned} \frac{x-2}{x+1}&=10 &&\text{quotient law and injectivity}\\ x-2&=10x+10 &&\text{multiply by }x+1>0\\ x&=-\frac43 &&\text{solve} \end{aligned}

Why this step? Solving the combined equation cannot override the domain of the two separate logarithms.

Answer. No real solution: $-4/3$ is outside $x>2$ .

Check. For $x>2$ , the quotient lies between zero and 1; it cannot equal 10.

The familiar sum $\ln(x-1)+\ln(x+2)=\ln4$ similarly gives $2,-3$ , with only 2 valid. An argument equal to zero is excluded too, even when another argument is positive.

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Pure Mathematics

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ME6 - Solving exponential and logarithmic equations

Taking logarithms to find an exponent

Equal bases and exponential substitutions

Logarithmic equations and original domains

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