How do you solve # 5^(x - 1) = 3^x#?

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Answer 1 · 2016-02-11T04:09:41

You must first convert to logarithmic form.

#log5^(x - 1) = log3^x#

#(x - 1)log5 = xlog3#

Distribute the parentheses:

#xlog5 - log5 = xlog3#

Put all x terms to the left side of the equation:

#xlog5 - xlog3 = log5#

Factor out the x:

#x(log5 - log3) = log5#

Use the quotient rule:

#x(log(5/3)) = log5#

#x = log5/(log(5/3)#

#x = log_(5/3)5#

Practice exercises:

a) #2^(x - 2) = 3^(2x + 4)#

b) #3^(3x) = 4 xx 5^(x - 6)#

Good luck!