How do you solve #4^x = 1/2#?

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Answer 1 · 2016-04-01T20:22:11

#x=-1/2#

Take the natural logarithm of both sides of the equation

#ln|4^x|=ln|1/2|#

Rewrite using properties of exponents

#xln|4|=ln|1/2|#

#xln|2^2|=ln|1|-ln|2|#

#2xln|2|=0-ln|2|#

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

#x=-1/2#

Answer 2 · 2016-04-03T01:56:51

Alternate answer:

Rewriting the exponents in a common base:

Don't forget that #1/(a^n) = a^-n#. Thus, #1/2 = 2^-1#.

#(2^2)^x = 2^-1#

Using the exponent property #(a^n)^m = a^(n xx m)#, we get the following:

#2^(2x) = 2^-1#

We can eliminate the bases now and solve the simple linear equation.

#2x = -1#

#x = -1/2#

Here are a few helpful exponent rules to know when working with harder problems:

#•a^n xx a^m = a^(n + m)#

#•a^n/a^m = a^(n - m)#

#•a^(n/m) = root(m)(a^n)#

Practice exercises:

#3^(2x + 1) xx 9^(x - 3) = 27^(4x - 5)#

#(4^(2x + 5))/(8^(3x - 2)) = 16^(2x)#

Challenge Problem:

Solve for x in the equation #sqrt(12) xx root(x)(12) = root(15)(12)#

Good luck!