How do you solve #5(1.034)^x-1998=13#?

1 Answer
Bill K.
Jul 23, 2015

#x=(ln(402.2))/(ln(1.034))\approx 179.363#

Explanation:

First, rearrange to write it as #5(1.034)^{x}=13+1998=2011#.

Next, divide both sides by #5# to get #1.034^{x}=2011/5=402.2#.

After that, take a logarithm of both sides (it doesn't matter what base you use). I'll use base #e#: #ln(1.034^{x})=ln(402.2)# or, by a property of logarithms, #x*ln(1.034)=ln(402.2)#.

Hence, #x=(ln(402.2))/(ln(1.034))\approx 179.363#.

You should check that this works by substitution back into the original equation.

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