1.06g - Solving exponential and logarithmic equations | OCR A-Level Mathematics A

Choose equal bases, logarithms or a substitution to solve exponential equations. Preserve logarithm domains when checking candidates, and recognise when a graph intersection gives a useful numerical answer instead of a simple algebraic one.

Taking logarithms to find an exponent

Will taking logarithms always free an unknown exponent? Compare $4\mathrm e^{2x-1}=7$ with $\mathrm e^x=3-x$ . The first can be untangled exactly; in the second, the unknown stays tangled inside another expression. We will learn where to change approach.

You already know

For positive $u$ , $\ln(u^r)=r\ln u$ . The identities $\ln(\mathrm e^v)=v$ and $\mathrm e^{\ln u}=u$ undo exponentials and logarithms. Keep brackets around a whole exponent.

For the first opening equation, isolate the exponential: $\mathrm e^{2x-1}=7/4$ , then $2x-1=\ln(7/4)$ . Thus $x=(1+\ln(7/4))/2\approx0.780$ , and exact substitution gives $4\mathrm e^{\ln(7/4)}=7$ .

Solving a positive-base exponential

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 follows by taking natural logarithms of both sides and using the power law.

Not in the formula sheet: learn it.

The denominator is nonzero because $a\ne1$ . This argument uses log laws, without requiring a change-of-base rule.

Worked example 1: a linear exponent

Question. Solve $5^{2x-1}=12$ , giving an exact answer.

Plan. Take natural logarithms and solve the resulting linear equation.

\begin{aligned} (2x-1)\ln5&=\ln12 &&\text{power law}\\ 2x-1&=\frac{\ln12}{\ln5} &&\text{divide by nonzero }\ln5\\ x&=\frac{1+\ln12/\ln5}{2} &&\text{isolate }x \end{aligned}

Why this step? The entire exponent multiplies $\ln5$ ; losing its brackets loses the constant term.

Answer. $x=\tfrac12(1+\ln12/\ln5)$ .

Check. The exponent is $\ln12/\ln5$ , so taking the logarithm of the resulting left side gives $\ln12$ , as required.

Change only the right side from a constant to an exponential involving $x$ .

Worked example 2: exponentials on both sides

Question. Solve $5^{2x-1}=12^x$ exactly.

Plan. Take logarithms of both positive sides, then collect the terms containing $x$ .

\begin{aligned} (2x-1)\ln5&=x\ln12 &&\text{power law on both sides}\\ x(2\ln5-\ln12)&=\ln5 &&\text{collect }x\text{ terms}\\ x&=\frac{\ln5}{\ln(25/12)} &&\text{solve the linear equation} \end{aligned}

Why this step? Different bases do not prevent a linear equation once both exponents become multipliers.

Answer. $x=\ln5/\ln(25/12)$ .

Check. With this value, $x(2\ln5-\ln12)-\ln5=0$ , so the two positive sides have equal logarithms and are equal.

The same route gives $2^{x+1}=3^{x-1}\Rightarrow x=\ln6/\ln(3/2)\approx4.42$ . Base-ten logarithms used consistently give the same answers. Keep unrounded values until the final line.

Watch out

For $b\le0$ , $a^x=b$ has no real solution when $a>0$ . For base 1, $1^x=b$ has every real input as a solution if $b=1$ , and none otherwise. Never divide by $\ln1=0$ .

Try the two-variable-exponent structure with fresh bases.

Equal bases and exponential substitutions

First look for a common base; then ask whether one exponential can stand in for the unknown.

For $3^{2x-1}=27=3^3$ , the one-to-one exponential gives $2x-1=3$ , so $x=2$ . Equal-exponent reasoning also works with $0<a<1$ , but not with base 1. When terms contain $\mathrm e^{2x}$ and $\mathrm e^x$ , the identity $\mathrm e^{2x}=(\mathrm e^x)^2$ creates a quadratic. For a general positive base, $a^{x+1}=a\,a^x$ may expose the same structure. These index identities are not supplied on the formula sheet.

Worked example 3: one admissible exponential value

Question. Solve $\mathrm e^{2x}-3\mathrm e^x-4=0$ .

Plan. Put $u=\mathrm e^x>0$ , solve the quadratic, then return to $x$ .

\begin{aligned} u^2-3u-4&=0 &&\text{substitute }u=\mathrm e^x\\ (u-4)(u+1)&=0 &&\text{factorise}\\ u&=4\text{ or }-1 &&\text{quadratic candidates}\\ \mathrm e^x&=4\quad\Rightarrow\quad x=\ln4 &&\text{retain only positive }u \end{aligned}

Why this step? The substitution has range $u>0$ , so $-1$ is rejected before attempting a logarithm.

Answer. $x=\ln4$ only.

Check. The starting left side becomes $16-12-4=0$ .

Change only the constant term, from $-4$ to 2; now both quadratic roots can survive.

Worked example 4: two admissible exponential values

Question. Solve $\mathrm e^{2x}-3\mathrm e^x+2=0$ .

Plan. Use the same positive substitution and check both roots.

\begin{aligned} u^2-3u+2&=(u-1)(u-2)=0 &&\text{factorise in }u\\ u&=1\text{ or }2 &&\text{both positive}\\ x&=\ln1=0\text{ or }\ln2 &&\text{return to }x \end{aligned}

Why this step? Reject roots using the range, not a habit of keeping only one answer.

Answer. $x=0$ or $x=\ln2$ .

Check. Substitution gives $1-3+2=0$ and $4-6+2=0$ .

A quadratic root is a value of $u$ , not yet a value of $x$ .

Exam tip

If asked to explain a rejection, say “ $\mathrm e^x>0$ for every real $x$ ”. A statement such as “negative logarithms are impossible” is false and does not justify the rejection.

Change the constant once more; the quadratic formula now gives exact surds.

Logarithmic equations and original domains

A log equation can produce algebraic candidates that do not belong to its starting domain.

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1.06g - 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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