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

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Answer 1 · 2016-10-29T22:03:39

How you do it is explained below.

Take the natural logarithm of both sides:

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

Use the identity #ln(a^b) = ln(a)b#

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

Distribute ln(5):

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

Subtract ln(5)x from both sides:

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

Multiply both sides by -1:

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

Divide both sides by #(ln(5)-ln(2))#:

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

#x ~~ 3.5#