How do you solve $10 {(1 + 0.25)}^{x} = 200$ $10(1+0.25)^x=200$ ?

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Answer 1 · 2018-07-17T12:06:06

$x \approx 13.425$ $x~~13.425$

$10 {(1 + 0.25)}^{x} = 200$ $10(1+0.25)^x=200$

Let's divide by 10 on both sides and simplify within the brackets:

${1.25}^{x} = 20$ $1.25^x=20$

Take the log on both sides:

$log ({1.25}^{x}) = log 20$ $log(1.25^x)=log20$

$x log 1.25 = log 20$ $xlog1.25=log20$

$x = \frac{log 20}{log 1.25} \approx 13.425$ $x=log20/log1.25~~13.425$

Let's check it:

$10 {(1 + 0.25)}^{13.425} = 199.994$ $10(1+0.25)^13.425=199.994$

Had we rounded with more significant digits, we would have gotten even closer to 200.

How do you solve 10(1+0.25)x=20010(1+0.25)^x=200?