How do you solve #8(10)^x + 2 = 26#?

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Answer 1 · 2018-06-24T14:10:29

#x=log_10(3)~~0.477#

First, rewrite to the from #a^x=b#. Then use #x=log_a(b)#.

#8(10)^x+2=26#

Subtract #2# from both sides.

#8(10)^x=24#

Divide both sides by #8#.

#10^x=3#

Take the common logarithm (#log# with base #10#) of both sides.

#log_10(10^x)=log_10(3)#

By definition: #log_a(a^x)=x#, so:

#x=log_10(3)~~0.477#