# Logarithm - Solve for X

• April 12th 2008, 07:10 PM
chrozer
Logarithm - Solve for X
How would you go by solving these problems to find the value of $x$?

1. $(log_3 x)(log_4 3) = 2$

2. $log x^3 - log y^2 =8$
$log x^4 + log y^6 = 2$
• April 12th 2008, 07:22 PM
Mathstud28
Quote:

Originally Posted by chrozer
How would you go by solving these problems to find the value of $x$?

1. $(log_3 x)(log_4 3) = 2$

2. $log x^3 - log y^2 =8$
$log x^4 + log y^6 = 2$

for one rewrite it as $log_3(x)+\frac{1}{log_3(4)}$...for the second one rewrite it as $log(x^4y^6)=2$
• April 12th 2008, 08:36 PM
o_O
1. Know this identity: $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ for any workable $c$

So:
$\log_{3}x \cdot \log_{4}{3} = 2$
$\frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2$

You should be able to simplify it form here.

2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

$\begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + & \log y^{6} & = & 2 \end{array}$ $. \quad \Rightarrow \quad .$ $\begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + & 6\log y & = & 2 \end{array}$

$\Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + & 6\log y & = & 2 & \mbox{ } \end{array}$

Add them and solve for x.
• April 12th 2008, 08:43 PM
Mathstud28
here is where I was going
Quote:

Originally Posted by o_O
1. Know this identity: $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ for any workable $c$

So:
$\log_{3}x \cdot \log_{4}{3} = 2$
$\frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2$

You should be able to simplify it form here.

2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

$\begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + & \log y^{6} & = & 2 \end{array}$ $. \quad \Rightarrow \quad .$ $\begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + & 6\log y & = & 2 \end{array}$

$\Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + & 6\log y & = & 2 & \mbox{ } \end{array}$

Add them and solve for x.

$\ln(x^4y^6)=2\Rightarrow{x^4y^6=e^2}$ and $\frac{y^2}{x^3}=e^8$ then solve...is that flawed logic?
• April 12th 2008, 08:58 PM
o_O
Ah I see now, going with a direct comparison.

And are we dealing with base 10 log or natural log?
• April 12th 2008, 09:03 PM
Mathstud28
Haha
Quote:

Originally Posted by o_O
Ah I see now, going with a direct comparison.

And are we dealing with base 10 log or natural log?

I never know anymore in math...I always taught that $log(x)\equiv{log_{10}(x)}$ but in most mathematical text $log(x)\equiv{ln(x)}$ so I just assume that I was taught wrong and use the latter...if he means $log_{10}(x)$ then just replace the e with 10
• April 12th 2008, 09:21 PM
chrozer
Quote:

Originally Posted by o_O
1. Know this identity: $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ for any workable $c$

So:
$\log_{3}x \cdot \log_{4}{3} = 2$
$\frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2$

You should be able to simplify it form here.

2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

$\begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + & \log y^{6} & = & 2 \end{array}$ $. \quad \Rightarrow \quad .$ $\begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + & 6\log y & = & 2 \end{array}$

$\Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + & 6\log y & = & 2 & \mbox{ } \end{array}$

Add them and solve for x.

I solved for "x" and got 100, but when I plugged it back into the equation it does not equal the answer.

EDIT - NVM....I'm an idiot. Plugged the "x" value for the "y" value also. But how would you solve for "y"? I can't get the same "y" value for both equation.

MathStud - Thanx for the help...but I never learned that $log(x) = ln(x)$
• April 12th 2008, 09:32 PM
Mathstud28
Ok
Quote:

Originally Posted by chrozer
I solved for "x" and got 100, but when I plugged it back into the equation it does not equal the answer.

MathStud - Thanx for the help...but I never learned that $log(x) = ln(x)$

well listing the adaptation I said we have $\frac{x^3}{y^2}=10^8\Rightarrow{x^3=10^8y^2}$...therefore using substitution we get $x=100,y=\frac{1}{10}$
• April 12th 2008, 09:53 PM
chrozer
Quote:

Originally Posted by Mathstud28
well listing the adaptation I said we have $\frac{x^3}{y^2}=10^8\Rightarrow{x^3=10^8y^2}$...therefore using substitution we get $x=100,y=\frac{1}{10}$

I got that while trying to solve for both "x" and "y" but when I plug it back into the equation on my calculator...it does not equal to the answer of the equation.
• April 12th 2008, 09:59 PM
o_O
$\log (x^{3}) - \log (y^{2}) = \log\left(100^{3}\right) - \log \left(.1^{3}\right) = 6 - (-2) = 8$

$\log (x^{4}) - \log (y^{6}) = \log\left(100^{4}\right) + \log \left(.1^{6}\right) = 8 + (-6) = 2$
• April 12th 2008, 10:01 PM
Mathstud28
Quote:

Originally Posted by chrozer
I got that while trying to solve for both "x" and "y" but when I plug it back into the equation on my calculator...it does not equal to the answer of the equation.

we have $log_{10}((10^2)^4)+log_{10}(10^{-6})=2\Rightarrow{8log_{10}(10)-6log_{10}(10)=2}$ $\Rightarrow{2log_{10}(10)=2}\Rightarrow{2=2}$ and $log_{10}((10^2)^3)-log_{10}(10^{-2})=8\Rightarrow{6log_{10}(10)+2log_{10}(10)=8}$ $\Rightarrow{8log_{10}(10)=8}\Rightarrow{8=8}$
• April 12th 2008, 10:05 PM
chrozer
Quote:

Originally Posted by o_O
$\log (x^{3}) - \log (y^{2}) = \log\left(100^{3}\right) - \log \left(.1^{3}\right) = 6 - (-2) = 8$

$\log (x^{4}) - \log (y^{6}) = \log\left(100^{4}\right) + \log \left(.1^{6}\right) = 8 + (-6) = 2$

Ok thnx alot. I see what I did wrong. I put the exponent outside the parentheses. Thnx all for your help.