• Dec 2nd 2009, 07:47 AM
bkbowser
$\displaystyle 2 log x - {2/3} log x + 4 log x = log x{16/3}$

I'm not confused by the logarithm operations. But apparently the fractional exponents. The answer given above is from the book. I thought it should be something like;
$\displaystyle 2*{3/2}*4={24/2}=12$
• Dec 2nd 2009, 08:20 AM
qmech
logs
$\displaystyle 2 log x - \frac{2}{3} log x + 4 log x$

$\displaystyle = log x^2 +log x ^ { -\frac {2}{3}} + log x^4$

$\displaystyle = log x^2 x ^ { -\frac {2}{3}} x^4$

$\displaystyle = log x^ { 2 -\frac {2}{3} + 4 }$

$\displaystyle = log x^{16/3}$

Alternatively:

$\displaystyle 2 log x - \frac{2}{3} log x + 4 log x$

$\displaystyle =(2 - \frac{2}{3} + 4) log x$

$\displaystyle =\frac{16}{3} log x$

$\displaystyle = log x^{16/3}$
• Dec 2nd 2009, 08:20 AM
Soroban
Hello, bkbowser!

Why are you multiplying and dividing exponents?

Quote:

$\displaystyle 2\log(x) - \tfrac{2}{3}\log(x) + 4\log(x) \:=\:\log \left(\frac{16}{3}\right)$

We have: .$\displaystyle \log\left(x^2\right) - \log\left(x^{\frac{2}{3}}\right) + \log\left(x^4\right) \:=\: \log\left(\frac{16}{3}\right)$

. . . . . . . $\displaystyle \log\left(\frac{x^2\cdot x^4}{x^{\frac{2}{3}}}\right) \:=\:\log\left(\frac{16}{3}\right)$

. . . . . . . $\displaystyle \log\left(x^{\frac{15}{3}}\right) \:=\:\log\left(\frac{16}{3}\right)$

"Un-log" both sides: .$\displaystyle x^{\frac{16}{3}} \:=\:\frac{16}{3}$

Therefore: .$\displaystyle x \:=\:\left(\frac{16}{3}\right)^{\frac{3}{16}}$

• Dec 2nd 2009, 08:46 AM
bkbowser
I thought you where supposed to multiply what ever came after the log notation when adding and divide when subtracting. The book never said what too do with exponents beyond to just apply them (2 log 2 = log 4). Since the notation is different enough I'm not really sure what too do most of the time. This gets made worse by the fact that the explanations of how to perform an operation don't include any information on why.

Oh and they where just asking to combine 3 log terms not solve for anything which it seems like you did to get an x = something statement. I was unclear on that.
• Dec 2nd 2009, 08:50 AM
qmech
This is not a question of "solve for x". The RHS of the original question is not
$\displaystyle log (16/3)$

$\displaystyle log (x)^{16/3}$
• Dec 2nd 2009, 09:03 AM
bkbowser
Quote:

Originally Posted by qmech
This is not a question of "solve for x". The RHS of the original question is not
$\displaystyle log (16/3)$

$\displaystyle log (x)^{16/3}$