How do you solve #2^x = 5^(x - 2)#?

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Answer 1 · 2016-07-15T17:30:11

#x = ( ln 25)/ (ln (5/2))#

take logs on each side

#ln 2^x = ln 5^(x - 2)#

then because #ln a^b = b ln a#

#x ln 2 = (x-2) ln 5#

#x (ln 5 - ln 2) = 2 ln 5#

#x = (2 ln 5)/ (ln 5 - ln 2)#

or if you like

#x = ( ln 25)/ (ln (5/2))#

because

#ln 5 - ln 2 = ln (5/2)#

and #2 ln 5 = ln 5^2 = ln 25#