# How do you solve 7^(2x-3) - 4 = 14 and find any extraneous solutions?

May 31, 2018

Solution: $x \approx 2.243$

#### Explanation:

${7}^{2 x - 3} - 4 = 14$ or

${7}^{2 x - 3} = 18$ Taking log on both sides we get,

$\left(2 x - 3\right) \log 7 = \log 18$ or

$\left(2 x - 3\right) = \log \frac{18}{\log} 7$ or

$\left(2 x - 3\right) = \log \frac{18}{\log} 7 \approx 1.485$ or

$2 x \approx 3 + 1.485 \approx 4.485$ or

$x \approx \frac{4.485}{2} \approx 2.243 \left(2 \mathrm{dp}\right)$

Check: ${7}^{2 \cdot 2.243 - 3} - 4 \approx 14$(verified)

No extraneous solution.

Solution: $x \approx 2.243 \left(3 \mathrm{dp}\right)$ [Ans]