Solve for x. 3^(x-1) = 2^(x-4)?

3ˣˉ¹ = 2ˣˉ⁴

Take the log base 3 of both sides:

log₃ 3ˣˉ¹ = log₃ 2ˣˉ⁴

Use the property of logs that says log Bˣ = xlogB:

x-1 = (x-4)log₃ 2

Distribute:

x-1 = xlog₃ 2 - 4log₃ 2

x-1 = xlog₃ 2 - log₃ 2⁴

x-1 = xlog₃ 2 - log₃ 16

Group x terms on L.H.S.:

x - xlog₃ 2 = 1 - log₃ 16

Factor out x to isolate it:

x(1-log₃ 2) = 1 - log₃ 16

x = (1 - log₃ 16) / (1-log₃ 2)

If you want to get fancy, getting it down to one single logarithm, you proceed:

x = (1 - log₃ 16) / (1-log₃ 2)

Substitute 1 = log₃3:

x = (log₃3 - log₃ 16) / (log₃3-log₃ 2)

Apply property of logs that says log A - log B = log(A/B):

x = (log₃ (3/16) / (log₃ 3/2)

Change of base property:

x = log{base 3/2} (3/16)

This link has a step by step solution to your problem:

http://www.symbolab.com/solver/equation-calculator...

Hope this helps