How do you solve $2^{x} = 7^{x - 1}$ $2^x=7^(x-1)$ ?

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How do you solve $2^{x} = 7^{x - 1}$ $2^x=7^(x-1)$ ?

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Answer 1 · 2016-05-30T07:13:23

$x = \frac{log (7)}{log (7) - log (2)} \approx 1.5533$ $x = log(7)/(log(7)-log(2)) ~~ 1.5533$

Explanation:

Dividing both sides by $7^{x}$ $7^x$ we find:

$\frac{1}{7} = \frac{2^{x}}{7^{x}} = {(\frac{2}{7})}^{x}$ $1/7 = 2^x/7^x = (2/7)^x$

Then taking logs (any base):

$log (\frac{1}{7}) = log ({(\frac{2}{7})}^{x}) = x log (\frac{2}{7})$ $log(1/7) = log((2/7)^x) = x log(2/7)$

Divide both sides by $log (\frac{2}{7})$ $log(2/7)$ and transpose to get:

$x = \frac{log (\frac{1}{7})}{log (\frac{2}{7})} = \frac{- log (7)}{log (2) - log (7)} = \frac{log (7)}{log (7) - log (2)} \approx 1.5533$ $x = log(1/7)/log(2/7) = (-log(7))/(log(2)-log(7)) = log(7)/(log(7)-log(2)) ~~ 1.5533$

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Complex solutions

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

$\frac{1}{7} = {(\frac{2}{7})}^{x} \cdot e^{2 k π i}$ $1/7 = (2/7)^x * e^(2kpii)$

$ln (\frac{1}{7}) = ln ({(\frac{2}{7})}^{x} \cdot e^{2 k π i}) = x ln (\frac{2}{7}) + 2 k π i$ $ln(1/7) = ln((2/7)^x * e^(2kpii)) = x ln(2/7) + 2kpii$

$x ln (\frac{2}{7}) = ln (\frac{1}{7}) - 2 k π i$ $x ln(2/7) = ln(1/7)-2kpii$

$x = \frac{ln (\frac{1}{7}) - 2 k π i}{ln (\frac{2}{7})} = \frac{- ln (7) - 2 k π i}{ln (2) - ln (7)} = \frac{ln (7) + 2 k π i}{ln (7) - ln (2)}$ $x = (ln(1/7)-2kpii)/ln(2/7) = (-ln(7)-2kpii)/(ln(2)-ln(7)) = (ln(7)+2kpii)/(ln(7)-ln(2))$

where $k$ $k$ is any integer.

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How do you solve $2^{x} = 7^{x - 1}$ $2^x=7^(x-1)$ ?

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