TMUA: Exponential equations
Solving exponential equations.
- Section 1 Part 1 · MM5 Exponentials and logarithms · spec MM5.3
- Papers 1 and 2
- 9 practice questions
- No calculator
What the specification covers
- a^x = b; equations reducible to it
- Quadratics in a^x: reject non-positive roots
Key ideas
- \(a^x=b\Rightarrow x=\log_ab\) (or take logs of both sides).
- Spot a quadratic in disguise: \(4^x=(2^x)^2\), \(5^{2x+1}=5\cdot(5^x)^2\).
- After solving for \(y=a^x\), reject non-positive values of \(y\): \(a^x>0\) always.
Common mistakes
- \(2^{x+1}\ne2^x+1\); it is \(2\cdot2^x\).
- A negative root for \(y\) gives no real \(x\).
Exam tip
Substitute \(y=a^x\) as soon as you see \(a^{2x}\) and \(a^x\) together.
Worked example
Worked example
What is the sum of the real solutions of \(4^x-6\cdot2^x+8=0\)?
- \(3\)
- \(1\)
- \(8\)
- \(2\)
- \(6\)
Answer: A: \(3\)
With \(y=2^x\): \(y^2-6y+8=0\), \(y=2\) or \(4\), so \(x=1\) or \(x=2\). The sum is 3.
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
What is the product of the real solutions of \(5^{2x+1}-26\cdot5^x+5=0\)?
- \(1\)
- \(-\frac{1}{5}\)
- \(-1\)
- \(5\)
- \(0\)
Show the answer and solution
Answer: C: \(-1\)
With \(y=5^x\): \(5y^2-26y+5=0\), \((5y-1)(y-5)=0\), \(y=\tfrac15\) or \(5\), so \(x=-1\) or \(1\). The product is \(-1\).
Question 2
The real number \(x\) satisfies \(2^{3x+1}=5^x\). Which of the following is equal to \(x\)? (All logarithms are to base 10.)
- \(\frac{\log 2}{\log 5-\log 3}\)
- \(\frac{\log 2}{\log 5-3\log 2}\)
- \(\frac{\log 2}{3\log 2-\log 5}\)
- \(\frac{1}{\log 5-3\log 2}\)
- \(\frac{\log 5}{3\log 2+1}\)
Show the answer and solution
Answer: B: \(\frac{\log 2}{\log 5-3\log 2}\)
Take logs: \((3x+1)\log2=x\log5\). So \(x(\log5-3\log2)=\log2\), \(x=\dfrac{\log2}{\log5-3\log2}\).
Question 3
What is the sum of the real solutions of \(5^{2x}-\frac{6}{25}\cdot5^x+\frac{1}{125}=0\)?
- \(2\)
- \(3\)
- \(-2\)
- \(\frac{1}{125}\)
- \(-3\)
Show the answer and solution
Answer: E: \(-3\)
With \(y=5^x\): \(y^2-\frac{6}{25}y+\frac{1}{125}=0\), whose roots are \(y=\frac{1}{25}\) and \(y=\frac{1}{5}\) (sum \(\frac{6}{25}\), product \(\frac{1}{125}\)). So \(x=-2\) or \(x=-1\), with sum \(-3\).
Question 4
What is the sum of the real solutions of \(3^{2x}-\frac{82}{3}\cdot3^x+9=0\)?
- \(2\)
- \(3\)
- \(-2\)
- \(-3\)
- \(9\)
Show the answer and solution
Answer: A: \(2\)
With \(y=3^x\): \(y^2-\frac{82}{3}y+9=0\), whose roots are \(y=27\) and \(y=\frac{1}{3}\) (sum \(\frac{82}{3}\), product \(9\)). So \(x=3\) or \(x=-1\), with sum \(2\).
Question 5
What is the sum of the real solutions of \(2^{2x}-\frac{3}{4}\cdot2^x+\frac{1}{8}=0\)?
- \(2\)
- \(-2\)
- \(-3\)
- \(\frac{1}{8}\)
- \(3\)
Show the answer and solution
Answer: C: \(-3\)
With \(y=2^x\): \(y^2-\frac{3}{4}y+\frac{1}{8}=0\), whose roots are \(y=\frac{1}{4}\) and \(y=\frac{1}{2}\) (sum \(\frac{3}{4}\), product \(\frac{1}{8}\)). So \(x=-2\) or \(x=-1\), with sum \(-3\).
Question 6
What is the sum of the real solutions of \(5^{2x}-\frac{626}{5}\cdot5^x+25=0\)?
- \(25\)
- \(3\)
- \(-3\)
- \(2\)
- \(-2\)
Show the answer and solution
Answer: D: \(2\)
With \(y=5^x\): \(y^2-\frac{626}{5}y+25=0\), whose roots are \(y=\frac{1}{5}\) and \(y=125\) (sum \(\frac{626}{5}\), product \(25\)). So \(x=-1\) or \(x=3\), with sum \(2\).
Question 7
What is the sum of the real solutions of \(2^{2x}-\frac{9}{4}\cdot2^x+\frac{1}{2}=0\)?
- \(\frac{1}{2}\)
- \(-2\)
- \(-1\)
- \(0\)
- \(1\)
Show the answer and solution
Answer: C: \(-1\)
With \(y=2^x\): \(y^2-\frac{9}{4}y+\frac{1}{2}=0\), whose roots are \(y=\frac{1}{4}\) and \(y=2\) (sum \(\frac{9}{4}\), product \(\frac{1}{2}\)). So \(x=-2\) or \(x=1\), with sum \(-1\).
Question 8
What is the sum of the real solutions of \(5^{2x}-\frac{126}{5}\cdot5^x+5=0\)?
- \(-2\)
- \(-1\)
- \(1\)
- \(5\)
- \(2\)
Show the answer and solution
Answer: C: \(1\)
With \(y=5^x\): \(y^2-\frac{126}{5}y+5=0\), whose roots are \(y=\frac{1}{5}\) and \(y=25\) (sum \(\frac{126}{5}\), product \(5\)). So \(x=-1\) or \(x=2\), with sum \(1\).
Keep going
- Previous topic: Exponential graphs and laws of logarithms
- Next topic: Differentiating powers of x
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