How do you solve #3log2x=4#?

1 Answer
Soumalya Pramanik · Anish M.
Mar 25, 2018

#x=(1/2)*10^(4/3)#

Explanation:

Assuming the logarithm as Common Logarithm (With base #10#),

#color(white)(xxx)3log2x=4#

#rArr log2x=4/3# [Transposing The 3 to R.H.S.]

#rArr 2x=10^(4/3)# [According to The Definition of Logarithm]

#rArr x=(1/2)*10^(4/3)# [Transposing 2 to R.H.S]

Hope this helps.

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