How do you solve #4^(3x-2)=1#?

2 Answers
Sep 10, 2016

Answer:

#x=2/3#

Explanation:

As any number to the power of zero is one, #=> 4^0=1#
Since #4^0=1 => 4^(3x-2)=1, => 3x-2=0#
Since #3x-2=0, => 3x=2, => x=2/3#

Sep 10, 2016

Answer:

To solve an exponential equation, take the log of each side. Or remember that anything raised to the zero power equals 1.

Explanation:

Take the log of each side.
#log4^(3x-2) =log1#

By log properties, the exponent becomes a coefficient
(#logx^a = alogx#)
#(3x-2)log4= log 1#

Divide both sides by log 4
#3x-2 = log1/log4#

#log1=0# so #log1/log4 = 0#

#3x-2=0#

Subtract 2 from each side.
#3x=2#

Divide each side by 3
#x=2/3#

OR

Remember that anything raised to the zero power equals 1
(#x^0=1#)

So the exponent #3x-2# must equal zero.

Solve
#3x-2=0#
#x=2/3#