Question

How do you solve `2^x=7^(x-1)`?

Answer 1

`x = log(7)/(log(7)-log(2)) ~~ 1.5533`

Explanation:

Dividing both sides by `7^x` we find:

`1/7 = 2^x/7^x = (2/7)^x`

Then taking logs (any base):

`log(1/7) = log((2/7)^x) = x log(2/7)`

Divide both sides by `log(2/7)` and transpose to get:

`x = log(1/7)/log(2/7) = (-log(7))/(log(2)-log(7)) = log(7)/(log(7)-log(2)) ~~ 1.5533`

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Complex solutions

We can find all Complex solutions by introducing a multiplier `e^(2kpii) = 1` during our calculations and using natural logarithms for convenience:

`1/7 = (2/7)^x * e^(2kpii)`

`ln(1/7) = ln((2/7)^x * e^(2kpii)) = x ln(2/7) + 2kpii`

`x ln(2/7) = ln(1/7)-2kpii`

`x = (ln(1/7)-2kpii)/ln(2/7) = (-ln(7)-2kpii)/(ln(2)-ln(7)) = (ln(7)+2kpii)/(ln(7)-ln(2))`

where `k` is any integer.