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

1 Answer
Jul 17, 2016

x=8

Explanation:

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