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

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Answer 1 · 2018-07-17T12:06:06

#x~~13.425#

#10(1+0.25)^x=200#

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

#1.25^x=20#

Take the log on both sides:

#log(1.25^x)=log20#

#xlog1.25=log20#

#x=log20/log1.25~~13.425#

Let's check it:

#10(1+0.25)^13.425=199.994#

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