Question

express as single logarithm and simplify if possible:
1/2log a^x+4log a y-2log a^x

Answer 1

I think it's log a sqrt x y^4

Answer 2

remember how I said the exponent can be brought out in front. Do that and see what you see

Answer 3

\[\large \log a^x= x \log a\]

Answer 4

sorry, logarithms are very hard for me to wrap my head around...

Answer 5

hmm, well, I have 4 possible answers, and none of them begin with x log a. I chose the one with the sqrt because of the 1/2

Answer 6

can u use the equation button and type out the question again

Answer 7

\[\frac{ 1 }{ 2 }\log_{a}x + 4\log_{a} y -2 \log_{a}x \] @ChmE does that help?

Answer 8

Much better. I couldn't tell that the "a" was a base in the question

Answer 9

oh yea, sorry

Answer 10

You can start by combining like terms. \[\frac{ 1 }{ 2 }\log_{a}x -2 \log_{a}x+ 4\log_{a} y=-\frac{ 3 }{ 2 } \log_{a}x+ 4\log_{a} y\]

Answer 11

then the coefficient can become the power of x and y

Answer 12

oh, ok. I had to think for a sec

Answer 13

ok, it's starting to look like a different answer: loga(y^4/x^3/2) so how do I get there from where we are now?

Answer 14

\[\log_{a} \left(\begin{matrix}y^{4} \\ x^{\frac{ 3 }{ 2 }}\end{matrix}\right)\]

Answer 15

\[\log_{a}x-\log_ay=\log_a {x \over y} \]

Answer 16

\[\log_{a}x+\log_ay=\log_a {x y}\]

Answer 17

Must have the same base for these rules to apply

Answer 18

oooh ok. I actually kinda see haha

Answer 19

Try to study the two rules I just posted. They are very important in logs. Also memorize how an exponent can be moved around when using logs. Good luck

Answer 20

thanks. I'll keep looking over this