Mathematics 9709 · Pure Mathematics 2

Unknown stuck in the power: a^x = b and the hidden quadratic

When the unknown is stuck up in a power (like $5^{2x+1}=7$ ), normal algebra can't reach it.
Take $\ln$ : the power law drops the exponent down to the front, where you can solve for it.

Unknown in a power → take $\ln$ of both sides. The power law turns $\ln(a^{\square})=\square\,\ln a$ . The exponent is now down at the front, in a normal (usually linear) equation.

The plain case: a^x = b

Worked example

Solve $5^{2x+1}=7$ , to 3 s.f.

Take $\ln$ of both sides: $\ln 5^{2x+1}=\ln 7$ .
Power law drops the exponent to the front: $(2x+1)\ln 5=\ln 7$ .
Now it's linear in $x$ . Divide, then unwrap: $2x+1=\dfrac{\ln 7}{\ln 5}\Rightarrow x=\dfrac{1}{2}\!\left(\dfrac{\ln 7}{\ln 5}-1\right)$ .

Answer

x=0.105\ \text{(3 s.f.)}

(textbook WE 2.11)

When the question asks for the answer in the form $\frac{\ln a}{\ln b}$ , leave the calculator alone. Keep everything as $\ln$ of whole numbers. Tidy it until it matches that shape. The exact form is the marks.

Worked example

Solve $2^{3x-1}=5\cdot 3^{-x}$ , in the form $\dfrac{\ln a}{\ln b}$

Take $\ln$ : $(3x-1)\ln 2=\ln 5-x\ln 3$ .
Expand and gather every $x$ -term on the left: $3x\ln 2+x\ln 3=\ln 5+\ln 2$ .
Factor and use the laws on each side: $x(\ln 8+\ln 3)=\ln 10\Rightarrow x\ln 24=\ln 10$ .
Divide: $x=\dfrac{\ln 10}{\ln 24}$ ( $\approx 0.725$ ).

Answer

x=\dfrac{\ln 10}{\ln 24}

(9709/31 Oct/Nov 2022 Q3)

Two different bases: 3^(2x) = 4^(x+5)

Same first move. Take $\ln$ , gather the $x$ -terms, then pull $x$ out:

$2x\ln 3=(x+5)\ln 4$ .
Expand and collect $x$ on one side: $2x\ln 3-x\ln 4=5\ln 4$ .
Factor out $x$ and divide: $x=\dfrac{5\ln 4}{2\ln 3-\ln 4}=8.55$ .

(textbook WE 2.11)

The hidden quadratic: spot the 2:1 powers

See two power terms whose exponents are in ratio 2 : 1 plus a number (like $3^{2x}$ and $3^{x}$ )? It is a hidden quadratic. Let $y=3^{x}$ , so $3^{2x}=(3^x)^2=y^2$ .

Worked example

Solve $2(3^{2x})+7(3^x)=15$ , to 3 s.f.

Substitute $y=3^x$ , so $3^{2x}=y^2$ : $2y^2+7y-15=0$ .
Calculator-first (or factorise $(2y-3)(y+5)=0$ ): $y=\tfrac32$ or $y=-5$ .
Reject $y=-5$ : $3^x$ is always positive, so $y$ can never be negative.
Back-substitute and take $\ln$ : $3^x=\tfrac32\Rightarrow x=\dfrac{\ln 1.5}{\ln 3}$ .

Answer

x=0.369\ \text{(3 s.f.)}

(textbook WE 2.12)

Common mistake

The trap: forgetting that

a^x>0

always. A negative

y

can never come from

3^x

, so drop it before you take logs. You can't take

\ln

of a negative number anyway.

The same method on an e^x quadratic

With base $e$ it looks harder, but it is the same: $e^{2x}-5e^x+6=0$ . Let $u=e^x$ : $u^2-5u+6=0\Rightarrow(u-2)(u-3)=0$ , so $e^x=2$ or $e^x=3$ , giving $x=\ln 2$ or $x=\ln 3$ . Here both roots survive because both are positive. So always check; don't just delete one. (textbook Exercise 2G Q11)

Your turn— tap to reveal the worked answer (9709/32 May/Jun 2022 Q1 (crossover with ln))

Solve $\ln(e^{2x}+3)=2x+\ln 3$ , to 3 d.p. Treat $e^{2x}$ as the unknown. Combine, then take $e$ of both sides: $\ln(e^{2x}+3)-2x=\ln 3\Rightarrow\ln\!\frac{e^{2x}+3}{e^{2x}}=\ln 3$ . So $e^{2x}+3=3e^{2x}\Rightarrow 2e^{2x}=3$ .

Answer

x=\tfrac12\ln\tfrac32=0.203\ \text{(3 d.p.)}

Inequalities: take logs, check the sign

Solving $a^x>b$ uses the same first move: take logs of both sides.
In the normal case $a>1$ the inequality stays the same way round.
You only flip it if the base is below $1$ (because then $\log_a$ gets smaller as the input grows). (textbook §2.6)

$2^x>3\Rightarrow x\ln 2>\ln 3\Rightarrow x>\dfrac{\ln 3}{\ln 2}$ . Just like an equation, but keep the $>$ when the base is above $1$ .

Common mistake

The flip you must never miss is dividing by a negative. Say gathering terms leaves

-x\ln 2>\ln 3

. Dividing by

-\ln 2

flips it to

x<-\frac{\ln 3}{\ln 2}

. The sign of the base and the sign of the number you divide by are two separate checks.

PreviousSolving ln / log equations (combine, then undo with e)Nexte^x and ln x: inverse twins, graphs, and the e/ln cancel

Unknown stuck in the power: a^x = b and the hidden quadratic

When the unknown is stuck up in a power (like $5^{2x+1}=7$ ), normal algebra can't reach it.
Take $\ln$ : the power law drops the exponent down to the front, where you can solve for it.

Unknown in a power → take $\ln$ of both sides. The power law turns $\ln(a^{\square})=\square\,\ln a$ . The exponent is now down at the front, in a normal (usually linear) equation.

The plain case: a^x = b

Worked example

Solve $5^{2x+1}=7$ , to 3 s.f.

Take $\ln$ of both sides: $\ln 5^{2x+1}=\ln 7$ .
Power law drops the exponent to the front: $(2x+1)\ln 5=\ln 7$ .
Now it's linear in $x$ . Divide, then unwrap: $2x+1=\dfrac{\ln 7}{\ln 5}\Rightarrow x=\dfrac{1}{2}\!\left(\dfrac{\ln 7}{\ln 5}-1\right)$ .

Answer

x=0.105\ \text{(3 s.f.)}

(textbook WE 2.11)

Worked example

Solve $2^{3x-1}=5\cdot 3^{-x}$ , in the form $\dfrac{\ln a}{\ln b}$

Take $\ln$ : $(3x-1)\ln 2=\ln 5-x\ln 3$ .
Expand and gather every $x$ -term on the left: $3x\ln 2+x\ln 3=\ln 5+\ln 2$ .
Factor and use the laws on each side: $x(\ln 8+\ln 3)=\ln 10\Rightarrow x\ln 24=\ln 10$ .
Divide: $x=\dfrac{\ln 10}{\ln 24}$ ( $\approx 0.725$ ).

Answer

x=\dfrac{\ln 10}{\ln 24}

(9709/31 Oct/Nov 2022 Q3)

Two different bases: 3^(2x) = 4^(x+5)

Same first move. Take $\ln$ , gather the $x$ -terms, then pull $x$ out:

$2x\ln 3=(x+5)\ln 4$ .
Expand and collect $x$ on one side: $2x\ln 3-x\ln 4=5\ln 4$ .
Factor out $x$ and divide: $x=\dfrac{5\ln 4}{2\ln 3-\ln 4}=8.55$ .

(textbook WE 2.11)

The hidden quadratic: spot the 2:1 powers

See two power terms whose exponents are in ratio 2 : 1 plus a number (like $3^{2x}$ and $3^{x}$ )? It is a hidden quadratic. Let $y=3^{x}$ , so $3^{2x}=(3^x)^2=y^2$ .

Worked example

Solve $2(3^{2x})+7(3^x)=15$ , to 3 s.f.

Substitute $y=3^x$ , so $3^{2x}=y^2$ : $2y^2+7y-15=0$ .
Calculator-first (or factorise $(2y-3)(y+5)=0$ ): $y=\tfrac32$ or $y=-5$ .
Reject $y=-5$ : $3^x$ is always positive, so $y$ can never be negative.
Back-substitute and take $\ln$ : $3^x=\tfrac32\Rightarrow x=\dfrac{\ln 1.5}{\ln 3}$ .

Answer

x=0.369\ \text{(3 s.f.)}

(textbook WE 2.12)

Common mistake

The trap: forgetting that

a^x>0

always. A negative

y

can never come from

3^x

, so drop it before you take logs. You can't take

\ln

of a negative number anyway.

The same method on an e^x quadratic

Your turn— tap to reveal the worked answer (9709/32 May/Jun 2022 Q1 (crossover with ln))

Answer

x=\tfrac12\ln\tfrac32=0.203\ \text{(3 d.p.)}

Inequalities: take logs, check the sign

Solving $a^x>b$ uses the same first move: take logs of both sides.
In the normal case $a>1$ the inequality stays the same way round.
You only flip it if the base is below $1$ (because then $\log_a$ gets smaller as the input grows). (textbook §2.6)

$2^x>3\Rightarrow x\ln 2>\ln 3\Rightarrow x>\dfrac{\ln 3}{\ln 2}$ . Just like an equation, but keep the $>$ when the base is above $1$ .

Common mistake

The flip you must never miss is dividing by a negative. Say gathering terms leaves

-x\ln 2>\ln 3

. Dividing by

-\ln 2

flips it to

x<-\frac{\ln 3}{\ln 2}

. The sign of the base and the sign of the number you divide by are two separate checks.