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

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

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Answer 1 · 2016-02-24T14:56:12

Convert to logarithmic form.

Explanation:

${log 7}^{x - 2} = {log 5}^{x}$ $log7^(x - 2) = log5^x$

$(x - 2) log 7 = x log 5$ $(x - 2)log7 = xlog5$

$x log 7 - 2 log 7 . = x log 5$ $xlog7 - 2log7.=x log 5$

$x log 7 - x log 5 = 2 log 7$ $xlog7 - xlog5 = 2log7$

$x (log 7 - log 5) = 2 log 7$ $x(log7 - log5) = 2log7$

$x = \frac{2 log 7}{log 7 - log 5}$ $x = (2log7)/(log7 - log5)$

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

$x = \frac{{log 7}^{2}}{log (\frac{7}{5})}$ $x = log7^2/log(7/5)$

$x = \frac{log 49}{log (\frac{7}{5})}$ $x = log49/log(7/5)$

Hopefully this helps!

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

1 Answer

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