How do you solve $7^(x - 2) = 5^x$ ?

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Answer 1 · 2016-02-24T14:56:12

Convert to logarithmic form.

$log7^(x - 2) = log5^x$

$(x - 2)log7 = xlog5$

$xlog7 - 2log7.=x log 5$

$xlog7 - xlog5 = 2log7$

$x(log7 - log5) = 2log7$

$x = (2log7)/(log7 - log5)$

You can simplify the answer further by using the rules $alogn = logn^a$ and $log_an - log_am = log_a(n / m)$

$x = log7^2/log(7/5)$

$x = log49/log(7/5)$

Hopefully this helps!

How do you solve 7^(x - 2) = 5^x7^(x - 2) = 5^x?