Question

log(x+1)=2

Answer 1

Take Antilog both the sides, you will get:
\[x + 1 = \log(2)\]

Now find x..

\[\log_{10}(2) = 0.3010\]

Answer 2

What's the base?

Answer 3

I assume 10^

Answer 4

I haven't learned about antilogs? I'm in pre-cal

Answer 5

\[\log_{10}10^{x}=x\]
Do you know the relation above?

Answer 6

We are dealing with logs to base 10. Taking antilogs of both sides we get:
x + 1 = 100
(The antilog of 2 is 100).

Answer 7

So x = 100 - 1

Answer 8

since you know for a fact that
\[\log_ab=x\] is the same as writing (in exponent form)
\[a^x=b\]
\[\log (x+1)=2\]
\[x+1=10^2\]
\[x=10^2-1\]
\[x=99\]

Answer 9

NB: