How do solve for x in a log question?

Question

Solve for x: #2log_"10"x+3 = 5log_"10"x#

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Answer 1 · 2018-02-23T09:07:16

#x=10#

Let's start by simplifying the equation.

#2log_10 x+3 = 5log_10 x#
#3 = 3log_10 x#
#1 = log_10 x#

Now, since #log_a b = c# is equivalent to #a^c = b#, we can change our equation to,

#10^1 = x#
#x = 10#
#square#