Question

7log y-2log x+ 1/2log z express as a single logarithm and simplify if possible

is the answer log y^7 (log z^1/2) ????
_________
log x^2

Answer 1

Not quite.
Your expression is in terms of three logs, not one.
We want everything under one log sign.
The three rules are\[a \log x=\log (x^a)\](You used this rule properly, by the way.)

But pay close attention to the next two.\[\log a+\log b=\log(ab)\]\[\log a-\log b=\log({a \over b})\]Notice how the log signs combine.

Here's what you did:\[\log a+\log b=(\log a)(\log b)\]and\[\log a-\log b={\log a \over \log b}\]These rules are wrong. Try again using the correct rules, your answer is very close!

Answer 2

is it log y^7 (log z^1/2)
_____
x^2

Answer 3

even closer! but we still have two log signs...
I will show you the answer to this one because you are so close, try to follow along and find your mistake...

Answer 4

\[7\log y-2\log x+ {1\over2}\log z=\log(y^7)-\log(x^2)+\log(z^{1/2})\]\[=\log({y^7\over x^2})+\log(z^{1/2})=\log({y^7z^{1/2}\over x^2})\]you just forgot to get the z into the log sign. Like I said, look for exactly where you made your mistakes.

Answer 5

wow thanks you made this much easier to see with your hints. i appreciate you taking the time to explain this to me. and I should of multiplied the z^1/2 yo just the y not the whole divisional problem:)

Answer 6

thanks for listening :)