How do you solve #2^(x+1) = 3^x#?

1 Answer
Feb 26, 2016

Convert to logarithmic form.

Explanation:

#log2^(x + 1) = log3^x#

Simplify using the rule #loga^n = nloga#

#(x + 1)log2 = xlog3#

Distribute on the left side. Don't forget you cannot multiply directly a log with a non-log (e.g #2 xx log3!= log6#but is equal to #2log3#)

#xlog2 + log2 = xlog3#

Put the x's to one side of the equation.

#log2 = xlog3 - xlog2#

#log2 = x(log3 - log2)#

Simplify the right side of the equation further by using the rule #logm - logn = log(m / n)#

#log2 = x(log(3/2))#

#log2/log(3/2) = x#

You will want to ask your teacher if he/she wants the answer in exact form or rounded off. Just make sure to check.

Practice exercises:

  1. Solve for x. Leave answers in exact form.

a) #2^(3x) = 5^(x + 1)#

b). #5^(x - 3) = 3^(2x + 1)#

Challenge problem

Find the value of x in #2^(4x - 6) = 5 xx 3^(x + 7)#