What is #x# in #8(10)^(x-3) + 9 = 81#?

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Answer 1 · 2015-11-08T22:41:34

#x = log_10(9)+3#

To solve for #x# in an exponent, we will need to use logarithms.
Logarithms have the useful property that
#ln(x^k) = kln(x)#

First, as usual, we will isolate the term containing #x#

#8(10)^(x-3) = 72#
#=> 10^(x-3) = 9#

Now, we will take the natural log of both sides.

#ln10^(x-3) = ln9#

Applying the property mentioned above gives us

#(x-3)ln(10) = ln9#

#=> x-3 = ln9/ln10#

#=> x = ln9/ln10 + 3#

or, using the property that #log_b(x)/ln_b(y) = log_y(x)#

#x = log_10(9)+3#