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How do you solve #log_5 6 +log_5 (2x^2)=log_5 48#?

1 Answer

A. S. Adikesavan

Oct 5, 2016

#x=+-2#

Explanation:

Use #log_b m+log_b n = log_b(mn)#

Here,# log_5 6+log_5(2x^2)= log_b (12x^2)=log_5 48#

Log function is a bijective (single valued) function. So,

#12x^2=48#, and so,

#x^2=4 to x =+-2#

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