How do you solve #2log_b4+log_b5-log_b10=log_bx#?

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Answer 1 · 2016-07-17T16:35:34

x=8

Since #nloga=loga^n# you have:

#log_b 4^2+log_b 5-log_b 10=log_b x#.

Since #logp+logq=log(pq)# you have:

#log_b (16*5)-log_b 10=log_b x#.

Since #logp-logq=log(p/q)# you have:

#log_b(80/10)=log_b x#.

Then #log_b8=log_b x#

so x=8