Question

log√3 (9) and log1/4 (16) can someone help

Answer 1

let's look at the first one:
\[\log_{\sqrt{3}} 9\]
If you want to know what's the result of it, we just have to apply the logarithm definition that states:
\[\log_{a} b=c <=>a ^{c}=b\]
Let's apply it:
\[\log_{\sqrt{3}} 9 <=> (\sqrt{3})^{x}=9\]
And I'll rewrite the sqrt of 3 and convert the 9 into 3^2:
\[(3^{\frac{ 1 }{ 2 }})^{x}=3^{2}\]
And since there the exponent of the exponent, I just multiply them:
\[3^{\frac{ x }{ 2 }}=3^{2}\]
The base of that exponential equation is the same on both sides, so I can work with the exponents only:
\[\frac{ x }{ 2 }=2\]
Solving for x we conclude that x=4.
So we have finished, we say that:
\[\log_{\sqrt{3}} 9=4\]

Answer 2

SWAG!!!

Answer 3

swag!

Answer 4

Awesome, thank you Owlcoffee.

Answer 5

Try applying that to the other one.

Answer 6

okay

Answer 7

I think it will be -2

Answer 8

that's correct.

Answer 9

thank's