Year 11 Methods Unit 1 — Using Logarithms to Solve Exponentials

Cambridge Methods 1&2, §13F  •  Tue 02 June 2026
📚 Also for this topic: 📄 Printable worksheet ✅ Solutions (answer key) ← Back: §13E Logarithms → Next: §13G Log graphs

Today's lesson

In §13D you solved $2^x=8$ by spotting $8=2^3$. But what about $2^x=5$? There's no clean integer that does it. Now that you know logarithms (§13E), you have the tool: take the log of both sides. By the end of the lesson you should be able to:

Learning intentions

Part 1 — The "take log of both sides" trick

The method. To solve $a^x=b$ (where $a,b>0$, $a\ne 1$, $b\ne 1$):
  1. Take $\log_{10}$ of both sides: $\log_{10}(a^x)=\log_{10}(b)$.
  2. Apply the power rule on the LHS: $x\,\log_{10}(a)=\log_{10}(b)$.
  3. Divide: $x=\dfrac{\log_{10}(b)}{\log_{10}(a)}$.
  4. Evaluate on the CAS for the decimal answer.
Note: any base works — $\ln$ (= $\log_e$) is just as good. The CAS will simplify both.

📺 Walkthrough: solve $2^x=5$ — take $\log_{10}$ of both sides, apply the power rule, divide, and decimal-approximate.

EXAMPLE 1 — solve $2^x=5$
Solve $2^x=5$, exact and to 2 decimal places.
  1. $\log_{10}(2^x)=\log_{10}(5)$.
  2. $x\log_{10}(2)=\log_{10}(5)$.
  3. $x=\dfrac{\log_{10}(5)}{\log_{10}(2)}$  (exact). On the CAS: $x\approx 2.3219$.
  4. $\boxed{x\approx 2.32}$ to 2 d.p.
EXAMPLE 2 — solve $7^{-x}=6$
Solve $7^{-x}=6$, exact and to 2 decimal places.
  1. $\log_{10}(7^{-x})=\log_{10}(6)$.
  2. $-x\log_{10}(7)=\log_{10}(6)\Rightarrow x=-\dfrac{\log_{10}(6)}{\log_{10}(7)}$.
  3. $x\approx -0.9208$, so $\boxed{x\approx -0.92}$.

CAS shortcut: solve(7^(-x)=6, x) gives the same exact value instantly.

EXAMPLE 3 — bracketed exponent
Solve $5^{2x-1}=7$, to 2 decimal places.
  1. $\log_{10}(5^{2x-1})=\log_{10}(7)$.
  2. $(2x-1)\log_{10}(5)=\log_{10}(7)$.
  3. $2x-1=\dfrac{\log_{10}(7)}{\log_{10}(5)}\approx 1.2091$.
  4. $2x\approx 2.2091\Rightarrow \boxed{x\approx 1.10}$.
EXAMPLE 4 — fractional base
Solve $0.3^{1-x}=12$, to 2 decimal places.
  1. $\log_{10}(0.3^{1-x})=\log_{10}(12)$.
  2. $(1-x)\log_{10}(0.3)=\log_{10}(12)$.
  3. $1-x=\dfrac{\log_{10}(12)}{\log_{10}(0.3)}$.
  4. Note $\log_{10}(0.3)$ is negative (because $0<0.3<1$), and $\log_{10}(12)$ is positive — so the RHS is negative.
  5. $1-x\approx -2.0639 \Rightarrow x\approx 3.0639 \Rightarrow \boxed{x\approx 3.06}$.

Practice 1.1 — solve for $x$, to 2 d.p.

  1. $3^x=10$
  2. $2^{x+1}=15$
  3. $4^{2x}=20$
  4. $0.5^x=7$
a) $x=\dfrac{\log 10}{\log 3}\approx 2.10$
b) $x+1=\dfrac{\log 15}{\log 2}\approx 3.907 \Rightarrow x\approx 2.91$
c) $2x=\dfrac{\log 20}{\log 4}\approx 2.161 \Rightarrow x\approx 1.08$
d) $x=\dfrac{\log 7}{\log 0.5}\approx -2.81$ (negative because base is between 0 and 1)

Part 2 — Exponential inequalities that need logs

Watch the sign flip! When you divide both sides of an inequality by $\log_{10}(b)$:
EXAMPLE 5 — base bigger than 1
Solve $4^x\le 10$, to 2 decimal places.
  1. $\log_{10}(4^x)\le \log_{10}(10)$  ($\log_{10}$ is increasing, so direction stays).
  2. $x\log_{10}(4)\le 1$.
  3. $\log_{10}(4)\approx 0.602>0$, so divide and keep direction: $x\le \dfrac{1}{\log_{10}(4)}\approx 1.6610$.
  4. $\boxed{x\le 1.66}$.
EXAMPLE 6 — base less than 1 (sign FLIP)
Solve $0.4^x < 5$, to 2 decimal places.
  1. $\log_{10}(0.4^x) < \log_{10}(5)$  (direction stays at this step — $\log_{10}$ is increasing).
  2. $x\log_{10}(0.4) < \log_{10}(5)$.
  3. $\log_{10}(0.4)\approx -0.398 < 0$, so dividing flips the inequality: $x > \dfrac{\log_{10}(5)}{\log_{10}(0.4)}\approx -1.7565$.
  4. $\boxed{x > -1.76}$.
  5. Sanity check: at $x=0$, $0.4^0=1<5$ ✓. At $x=-2$, $0.4^{-2}=6.25\not<5$. The break is between $-2$ and $0$ — consistent with $x>-1.76$.

Practice 2.1 — solve the inequality (2 d.p.).

  1. $5^x > 12$
  2. $2^x \le 100$
  3. $0.7^x > 3$
a) $x>\dfrac{\log 12}{\log 5}\approx 1.54$
b) $x\le \dfrac{\log 100}{\log 2}\approx 6.64$
c) $\log 0.7<0$ ⇒ flip: $x<\dfrac{\log 3}{\log 0.7}\approx -3.08$

Part 3 — Finding the $x$-intercept of an exponential graph

Back in §13C you sketched things like $f(x)=2\times 10^x-4$ and identified the asymptote and $y$-intercept — but the $x$-intercept needed logs. Now you can finish the job.

EXAMPLE 7 — sketch $f(x)=2\times 10^x-4$
Sketch $f(x)=2\times 10^x-4$, giving the equation of the asymptote and both axis intercepts.
  1. Asymptote: as $x\to -\infty$, $10^x\to 0$, so $f(x)\to -4$. Asymptote $y=-4$.
  2. $y$-intercept: $f(0)=2(1)-4=-2$, so $(0,-2)$.
  3. $x$-intercept: set $2\times 10^x-4=0$ ⇒ $10^x=2$ ⇒ $x=\log_{10}(2)\approx 0.301$.
  4. Shape: increasing (since $10^x$ is increasing and coefficient $>0$), passing through $(0,-2)$ and $(\log_{10}(2),0)$, approaching $y=-4$ from above on the left.
  5. Sketch key points: asymptote $y=-4$ (dashed), $y$-int $(0,-2)$, $x$-int $(\log_{10}(2),0)\approx (0.30,0)$.

Part 4 — Quick quiz (5 min)

Pick the correct answer for each, then click Mark.

Q1. The exact solution of $3^x=10$ is

Q2. $5^{2x-1}=7$ to 2 d.p. gives

Q3. Solve $4^x\le 10$ (2 d.p.).

Q4. Solve $0.4^x < 5$ (2 d.p.).

Q5. The $x$-intercept of $f(x)=2\times 10^x-4$ is at

Working program — Cambridge Ex 13F

After the quiz, open Cambridge Chapter 13 and work through Exercise 13F:

QuestionFoundationStandard (set work)Advanced
Q1 — basic $a^x=b$1 (a,c,e)1*1*
Q2 — linear exponent2 (a)2*2*
Q3 — fractional base—3*3*
Q4 — inequalities4 (a,c)4*4*
Q5 — graph $x$-intercepts—5*5 + extension

Next lesson (13G): graphs of logarithmic functions and inverses.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:
  1. What is the very first step when solving $a^x=b$ algebraically?
  2. When solving an exponential inequality, when does the inequality direction flip?
  3. If $10^x=2$, what does $x$ equal exactly (no decimals)?

— Mr Wong