How do you solve #3^(x+2) = 10^(x-1)#?

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Answer 1 · 2018-07-13T13:36:35

I tried this:

I would take the logarithm in base #10# (written #"L"og#) of both sides:

#"L"og(3^(x+2))="L"og(10^(x-1))#

use a property of log to manipulate the powers:

#(x+2)"L"og(3)=(x-1)"L"og(10)#

but #"L"og(10)=1# so we get:

#(x+2)"L"og(3)=x-1#

rearrange:

#x"L"og(3)+2"L"og(3)-x=-1#

#x["L"og(3)-1]=-2"L"og(3)-1#

#x=(-2"L"og(3)-1)/["L"og(3)-1]=3.73746#

using a normal pocket calculator to evaluate the log in base #10#.