Question

Whats the answer to log(9x+5)=2+log(x^2-1)

Answer 1

(Taking base of each log as 10)

We can write 2 as : \(\log (10)^2\)

So we get : \(\log(9x+5) = \log(10)^2 + \log(x^2-1) \)

\(\implies \log(9x+5) - \log(10)^2 = \log(x^2-1) \)

Answer 2

Now, we can use the identity : \(\log a - \log b = \log{\cfrac{a}{b}}\)

So, we get in LHS :

\(\implies \log({\cfrac{9x+5}{10^2}}) = \log(x^2-1)\)

Answer 3

Now, just take antilog both sides and you will get :

\(\cfrac{9x+5}{10^2} = x^2-1\)

Answer 4

Now, you can easily solve for x, is everything clear to you?