Question

how do u solve, log6(a^2)+2)+log2^2=2

Answer 1

Can you rewrite ur question?
it is not clear

Answer 2

log 6^ (a^2)+2+log6^(2)=2

Answer 3

log 6^[(a^2)+2]+log6^(2)=2

Answer 4

oh that makes sense. Wait let me solve it

Answer 5

Wait is the 6 meant to be subscript?

Answer 6

Is six the base or is it a base of 10?

Answer 7

Do you mean, solve

\[\log_{6} (a ^{2}+2)+\log_{6}2=2\]?

Answer 8

6 is the base, @lokisan yes!!:)

Answer 9

So the answer is a = 4

Answer 10

no i wrote that before and It came out incorrect

Answer 11

6^2=2(a^2+2)
36=2(a^2+2)
18=a^2+2
16=a^2
4=a

Answer 12

OK.

\[\log_{6}(a^{2}+2)+\log_{6}2=\log_{6}[(a^{2}+2)(2)]=2\]
so, by definition of the logarithm,
\[(a^{2}+2)2=6^{2} {\rightarrow}a^{2}=16 {\rightarrow}a={\pm}4\]

Answer 13

Oh ya. I forgot about that

Answer 14

Thanks!!

Answer 15

You're welcome.