- anonymous

If (log_3 X) (log_5 3) = 3, find x

- katieb

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- anonymous

Which one of these has an x?

- anonymous

The other one is just a multiplicative constant. You can divide it over to the other side.

- anonymous

?? how? the bases are diff-- three and five...how can u just divide?

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## More answers

- anonymous

Because the log base 5 of 3 is just a number. Like 4/5 or the square root of 15

- anonymous

=( i still don't get it....

- anonymous

Ok here. Lets say that \[C=log_5 3\]
Now your equation looks like:
\[C(log_3 x) = 3\]
Can you solve it now?

- anonymous

hmm..you know how log_b Y = x is equal the y=b^x in exponential form? how could you do that with this equation? isn't that how ur supposed to solve these things?

- anonymous

Yes, but first you have to get \(log_3x\) by itself.

- anonymous

so it's log_3 x = 3c and then 3^3c=x

- anonymous

Nope

- anonymous

whaat???nooo

- anonymous

C(log_3 x) = 3
So you have to divide by C, not multiply

- anonymous

OH. lol...right.

- anonymous

when i solve for (log_5 3)--to conv to exp form, does it still equal to three, which is the what the whole equation is equal to?

- anonymous

No.

- anonymous

That part is trickier

- anonymous

But lets see what you have with C for now.

- anonymous

x=3^3/c

- anonymous

Is that (3^3)/c or 3^(3/c) ?

- anonymous

second one

- anonymous

Correct.
Ok, so now to find what C is.
\[C = log_5 3 \iff 5^C = 3\]

- anonymous

With me so far?

- anonymous

yep

- anonymous

Ok, so that means that if we take the ln of both sides we get
\[C(ln\ 5) = (ln\ 3)\]

- anonymous

polpak i need ur help please

- anonymous

right?

- anonymous

yep

- anonymous

So that means \[C = \frac{(ln\ 3)}{(ln\ 5)}\]

- anonymous

oh~~~

- anonymous

Which means that
\[log_53 = \frac{(ln\ 3)}{(ln\ 5)}\]

- anonymous

oh, wow! thanks so much!!

- anonymous

that makes so much more sense now.

- anonymous

Glad I could help.

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