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

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Answer 1 · 2018-05-31T12:22:52

Solution: $x \approx 2.243$ $x~~ 2.243$

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

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

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

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

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

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

$x \approx \frac{4.485}{2} \approx 2.243 (2 d p)$ $x ~~ 4.485/2 ~~ 2.243(2 dp)$

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

No extraneous solution.

Solution: $x \approx 2.243 (3 d p)$ $x~~ 2.243 (3 dp)$ [Ans]

How do you solve 72x−3−4=147^(2x-3) - 4 = 14 and find any extraneous solutions?