Addition/Subtraction of logarithms

**bkbowser** · December 2nd 2009, 07:47 AM

$2 log x - {2/3} log x + 4 log x = log x{16/3}$

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;
$2*{3/2}*4={24/2}=12$

**qmech** · December 2nd 2009, 08:20 AM

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

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

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

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

$ = log x^{16/3} $

Alternatively:

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

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

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

$ = log x^{16/3} $

**Soroban** · December 2nd 2009, 08:20 AM

Hello, bkbowser!

Why are you multiplying and dividing exponents?

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

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

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

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

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

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

**bkbowser** · December 2nd 2009, 08:46 AM

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.

**qmech** · December 2nd 2009, 08:50 AM

This is not a question of "solve for x". The RHS of the original question is not
$ log (16/3) $

Instead, it is:

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

**bkbowser** · December 2nd 2009, 09:03 AM

Originally Posted by qmech

This is not a question of "solve for x". The RHS of the original question is not
$ log (16/3) $

Instead, it is:

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

What is "RHS"?

**qmech** · December 2nd 2009, 09:37 AM

RHS = right hand side
LHS = left hand side

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