What is 6.5: Solving exponential equations?

For a>0, a1 and b>0, taking logarithms gives a^x=b x=_a b= b/ a.

How many marks are 6.5 questions worth?

6.5 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 6.5 appeared in?

We have not yet indexed a verified past-paper appearance for 6.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.5

Solve equations of the form aˣ = b.

Draft — not yet indexed

Solving exponential equations

Worked answers and methods for 6.5 on Edexcel A-level Maths 9MA0.

Explanation

For $a>0$ , $a\neq1$ and $b>0$ , taking logarithms gives $a^x=b\Rightarrow x=\log_a b=\ln b/\ln a$ .
If the exponent is linear, isolate the exponential or take logarithms first, then solve the resulting linear equation without rounding intermediate values.
When both $a^x$ and $a^{-x}$ occur, substitute $u=a^x>0$ , so $a^{-x}=1/u$ , and solve the resulting algebraic equation before converting back.
A positive-base exponential cannot equal zero or a negative number; after a substitution, reject non-positive values because $a^x$ is always positive.
Substitute the final value into the original equation to verify it.

Worked example

Find the solution of $3^{2x-1}=20$ , giving $x$ to $3$ decimal places.

1.Taking logarithms gives $(2x-1)\ln3=\ln20$ .
2.Therefore $2x-1=\ln20/\ln3$ and $x=\tfrac12(1+\ln20/\ln3)=1.8634\ldots$ , so $x=1.863$ .

Answer: $x=1.863$

Common mistakes

Don't take logarithms but forget to multiply the exponent by $\log a$ .
Don't accept a non-positive substituted value for $a^x$ , even though a positive-base exponential is always positive.

Exam tip

Use one logarithm base consistently and keep full calculator precision until the requested final rounding.

Worked practice

Tier 1 · Easy

Solve

5^x=17

, giving

x

3

decimal places.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$x=1.760$	2
Notes
Take natural logarithms: $x\ln5=\ln17$ , so $x=\ln17/\ln5=1.7603\ldots$ , which gives $1.760$ .

(2 marks)

Tier 2 · Standard

Solve

5^{2x-1}=40

, giving your answer exactly in logarithmic form and to

3

decimal places.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$x=\dfrac12\left(1+\dfrac{\ln40}{\ln5}\right)$ , or any equivalent exact form: $\tfrac12\left(1+\log_5 40\right)$ , $\tfrac12\log_5 200$ and $\log_5\sqrt{200}$ are all accepted, in any base. $x=1.646$ to $3$ decimal places	3
Notes
Taking natural logarithms gives $(2x-1)\ln5=\ln40$ . Hence $2x-1=\ln40/\ln5$ , so $x=\frac12(1+\ln40/\ln5)=1.646014\ldots$ .

(3 marks)

Tier 3 · Hard

Determine all real solutions of

2^x+2^{-x}=5/2

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=-1$ or $x=1$	5
Notes
Let $u=2^x$ , so $u>0$ and $2^{-x}=1/u$ . Then $u+1/u=5/2$ . Multiplying by $2u$ gives $2u^2-5u+2=0=(2u-1)(u-2)$ . Thus $u=1/2$ or $u=2$ , so $2^x=2^{-1}$ or $2^x=2^1$ , giving $x=-1$ or $x=1$ .

(5 marks)

Tier 1 · Easy

Solve

7^{x+1}=19

, giving your answer to

3

decimal places.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x=0.513$	2
Notes
Taking logarithms gives $(x+1)\ln7=\ln19$ . Hence $x=\ln19/\ln7-1=0.513142\ldots$ , so $x=0.513$ to $3$ decimal places.

(2 marks)

Tier 2 · Standard

Solve

2^{x+1}=5^{2-x}

. Give your answer exactly in logarithmic form and to

3

decimal places.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$x=\dfrac{2\ln5-\ln2}{\ln10}$ $x=1.097$ to $3$ decimal places	4
Notes
Taking logarithms gives $(x+1)\ln2=(2-x)\ln5$ . Hence $x(\ln2+\ln5)=2\ln5-\ln2$ , so $x=(2\ln5-\ln2)/\ln10=1.0969\ldots$ .

(4 marks)

Tier 3 · Hard

Find all real solutions of

9^x-9\cdot3^x+14=0

, giving exact answers.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x=\log_3 2$ or $x=\log_3 7$	5
Notes
Let $u=3^x$ , so $u>0$ and $9^x=u^2$ . The equation becomes $u^2-9u+14=0=(u-2)(u-7)$ . Thus $3^x=2$ or $3^x=7$ , giving $x=\log_3 2$ or $x=\log_3 7$ .

(5 marks)

Tier 2 · Standard

The equation

5^{2x+k}=13

has the solution

x=1.2

. Find the value of

k

, giving your answer to three decimal places.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$k=-0.806$	3
Notes
Substituting $x=1.2$ gives $5^{2.4+k}=13$ . Taking logarithms gives $(2.4+k)\ln5=\ln13$ , so $k=\ln13/\ln5-2.4=-0.806307\ldots$ . Therefore $k=-0.806$ to three decimal places.

(3 marks)

Tier 3 · Hard

The equation

4^x-k\,2^x+9=0

, where

k

is a real constant, has exactly one real solution. Find

k

and the exact solution for

x

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$k=6$ $x=\log_2 3$	6
Notes
Let $u=2^x$ , so $u>0$ and $4^x=u^2$ . The equation becomes $u^2-ku+9=0$ . If its discriminant is negative, there is no real value of $u$ and hence no real solution for $x$ . If its discriminant is positive, the two roots have product $9>0$ and therefore have the same sign: for $k>0$ both are positive and give two values of $x$ , while for $k<0$ both are negative and give none. Exactly one positive root must therefore be repeated, so $k^2-36=0$ and $k=\pm6$ . If $k=-6$ , the repeated root is $u=-3$ , which is impossible. Hence $k=6$ , giving $(u-3)^2=0$ . Thus $2^x=3$ and $x=\log_2 3$ .

(6 marks)

Tier 3 · Hard

Solve

5^{2x-1}>7^{x+2}

, giving the exact boundary for

x

. Hence find the smallest integer value of

x

satisfying the inequality.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$x>\dfrac{\ln5+2\ln7}{2\ln5-\ln7}$ Smallest integer $x=5$	5
Notes
Taking logarithms gives $(2x-1)\ln5>(x+2)\ln7$ . Hence $x(2\ln5-\ln7)>\ln5+2\ln7$ . The coefficient is positive because $25>7$ , so division preserves the inequality and gives the stated boundary. This boundary is $4.32\ldots$ , so the smallest integer solution is $5$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Solve simultaneously for the real numbers

x

and

y

, given that

2^x3^y=12

and

\dfrac{4^x}{3^y}=\dfrac{16}{3}

. Give the exact values of

x

and

y

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$x=2$ and $y=1$	4
Notes
Multiplying the equations eliminates $3^y$ and gives $8^x=64=8^2$ , so $x=2$ . Substitution into $2^x3^y=12$ gives $4(3^y)=12$ , hence $3^y=3$ and $y=1$ . These values satisfy both original equations.

(4 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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