How do you solve #\log _ { 10} ( x - \pi ) = 5#?

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Answer 1 · 2017-03-31T20:33:09

#x=100000+pi#

Okay, so we know that the base of a log is also the base of an exponent.

#a=x^b# is #log_xa=b#

Knowing this, let's write #log_10(x-pi)=5# in exponential form.

#10^5=x-pi#

Simplify

#100000=x-pi#

Solve for #x#

#100000+pi=x#

#rArrx=100000+pi#