How do you solve #5000(1.09^(12x)) = 9500#?

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Answer 1 · 2016-01-07T00:21:07

#x=0.62066899187#

Divide both sides by #5000#.

#1.09^(12x)=1.9#

To get the #x# out of the exponent, or to undo the exponential function, take the logarithm with base #1.09# of both sides.

#log_1.09(1.09^(12x))=log_1.09 1.9#

#12x=log_1.09 1.9#

Divide both sides by #12#.

#x=(log_1.09 1.9)/12#

To find the value of this in a calculator, use the change of base formula, which states that

#log_ab=log_cb/log_ca#

Thus

#log_1.09 1.9=ln1.9/ln1.09#

so

#x=(ln1.9/ln1.09)(1/12)=ln1.9/(12ln1.09)=0.62066899187#