one logarithm problem

[starr22]

log[sub:3vlepkvh]3[/sub:3vlepkvh]x^2=2log[sub:3vlepkvh]3[/sub:3vlepkvh]4-4log[sub:3vlepkvh]3[/sub:3vlepkvh]5


please help me!!!

 

[galactus]

We could convert to base 10 and proceed from there.

\(\displaystyle \frac{log_{3}x^{2}}{log(3)}=\frac{log(16)-log(625)}{log(3)}\)

\(\displaystyle log x^{2}=log(\frac{16}{625})\)

\(\displaystyle x^{2}=\frac{16}{625}\)

\(\displaystyle x=\frac{4}{25}\)

 

[Mrspi]

log[sub:1x3dtijt]3[/sub:1x3dtijt]x^2=2log[sub:1x3dtijt]3[/sub:1x3dtijt]4-4log[sub:1x3dtijt]3[/sub:1x3dtijt]5


please help me!!!

Or, we could stay with log[sub:1x3dtijt]3[/sub:1x3dtijt]. Rewrite the right side as a single log, using the rules of logs.

THANK YOU, Galactus, for pointing out my typo!

a log b = log b[sup:1x3dtijt]a[/sup:1x3dtijt]

log[sub:1x3dtijt]3[/sub:1x3dtijt] x[sup:1x3dtijt]2[/sup:1x3dtijt] = log[sub:1x3dtijt]3[/sub:1x3dtijt] 4[sup:1x3dtijt]2[/sup:1x3dtijt] - log[sub:1x3dtijt]3[/sub:1x3dtijt] 5[sup:1x3dtijt]4[/sup:1x3dtijt]
log[sub:1x3dtijt]3[/sub:1x3dtijt] x[sup:1x3dtijt]2[/sup:1x3dtijt] = log[sub:1x3dtijt]3[/sub:1x3dtijt] 16 - log[sub:1x3dtijt]3[/sub:1x3dtijt] 625

llog a - log b = log (a/b)

log[sub:1x3dtijt]3[/sub:1x3dtijt] x[sup:1x3dtijt]2[/sup:1x3dtijt] = log[sub:1x3dtijt]3[/sub:1x3dtijt] (16/625)

Now...the bases are the same on both sides, and the two expressions are equal, so the arguments must be equal. If log[sub:1x3dtijt]b[/sub:1x3dtijt] m = log[sub:1x3dtijt]b[/sub:1x3dtijt] n, then m = n.

So, x[sup:1x3dtijt]2[/sup:1x3dtijt] = 16/625

Now finish it......

 

[galactus]

Be careful, Mrs P, that should be \(\displaystyle 2log_{3}(4)\), not \(\displaystyle 2log_{3}(3)\).

Just pointing out the typo. I do that often myself

 

[starr22]

THANKS GUYSSS!

 

[lookagain]

We could convert to base 10 and proceed from there.

\(\displaystyle (*) \rightarrow \frac{log(x^{2})}{log(3)} = \frac{log(16)-log(625)}{log(3)}\)

\(\displaystyle log x^{2} = log(\frac{16}{625})\)

\(\displaystyle x^{2} = \frac{16}{625}\)

\(\displaystyle (**) \rightarrow x = \pm \frac{4}{25}\)

In \(\displaystyle (*)\) there is no base \(\displaystyle 3\) in the numerator of the first fraction,
because you changed it all over to base \(\displaystyle 10\).

In \(\displaystyle (**)\) you get a positive or a negative value, and both of them
check, because you are putting them into \(\displaystyle x^2\) before the logarithm
is taken. Here, you are never taking the log of zero or a negative number.