How do you solve #2log_8(2x-5)=4#?

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Answer 1 · 2015-11-25T23:52:01

#x = 69/2#

First of all, your domain is #2x - 5 > 0 <=> x > 5/2# since the argument of any logarithmic expression needs to be greater than zero.

Now, to solve the equation, you should first divide both sides of the equation by #2#:

#color(white)(xx)2 log_8(2x-5) = 4#
#<=> log_8(2x-5) = 2#

The inverse function of #log_8(x)# is #8^x#. This means that #log_8(8^x) = x# and #8^(log_8 x) = x#.

In other words, you can make #log_8# disappear by applying the exponential function #8^x# on both sides!

#<=> 8^(log_8(2x-5)) = 8^2#

#<=> 2x - 5 = 64#

#<=> 2x = 69#

#<=> x = 69/2#