simple log question

**porge111** · July 21st 2010, 05:38 AM

"solve the following equation for x"

4^(3x-2) = 26^(x+1)

i use logs and get to :

(3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

thanks

**HallsofIvy** · July 21st 2010, 05:49 AM

Originally Posted by porge111

"solve the following equation for x"

4^(3x-2) = 26^(x+1)

i use logs and get to :

(3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

thanks

"log 4" and "log 26" are just numbers. If it were (3x- 2)A= (x+ 1)B for A and B any numbers, you would multiply to get 3Ax- 2A= Bx+ B. Subtract Bx from both sides and add 2A to both sides to get (3A- B)x= B+ 2A. Finally, (as long as $3A\ne B$ ) divide both sides by 3A- B to get $x= \frac{B+ 2A}{3A- B}$ . Now, replace A by log 4 and replace B by log 26.

**Prove It** · July 21st 2010, 05:58 AM

$4^{3x - 2} = 26^{x + 1}$

$\ln{\left(4^{3x - 2}\right)} = \ln{\left(26^{x + 1}\right)}$

$(3x - 2)\ln{4} = (x + 1)\ln{(26)}$

$(3x - 2)\ln{(2^2)} = (x + 1)\ln{(26)}$

$2(3x - 2)\ln{(2)} = (x+1)\ln{(26)}$

$\frac{2(3x - 2)}{x + 1} = \frac{\ln{(26)}}{\ln{(2)}}$

$\frac{6x - 4}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}$

$\frac{6x+6-10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}$

$\frac{6x+6}{x+1} - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}$

$\frac{6(x+1)}{x+1} - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}$

$6 - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}$

$-\frac{10}{x + 1} = \frac{\ln{(26)}}{\ln{(2)}} - 6$

$\frac{10}{x+1} = \frac{6\ln{(2)}}{\ln{(2)}} - \frac{\ln{(26)}}{\ln{(2)}}$

$\frac{10}{x + 1} = \frac{6\ln{(2)} - \ln{(26)}}{\ln{(2)}}$

$\frac{x + 1}{10} = \frac{\ln{(2)}}{6\ln{(2)} - \ln{(26)}}$

$x + 1 = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}}$

$x = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}} - 1$

$x = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}} - \left(\frac{6\ln{(2)} - \ln{(26)}}{6\ln{(2)} - \ln{(26)}}\right)$

$x = \frac{10\ln{(2)} + \ln{(26)} - 6\ln{(2)}}{6\ln{(2)} - \ln{(26)}}$

$x = \frac{4\ln{(2)} + \ln{(26)}}{6\ln{(2)} - \ln{(26)}}$ .

**Soroban** · July 21st 2010, 11:41 AM

Hello, porge111!

Solve for $x\!:\;\;4^{3x-2} \:=\: 26^{x+1}$

i used logs and get to: . $(3x-2)\log4 \:=\:(x+1)\log26$

... but not sure where to go from here.

You have: . $3x\!\cdot\!\log4 - 2\!\cdot\!\log 4 \:=\:x\!\cdot\!\log26 + \log26$

. . . . . . . $3x\!\cdot\!\log4 - x\!\cdot\!\log26 \;=\;2\!\cdot\!\log4 + \log26$

Factor: . $x\!\cdot\!\bigg(3\log4 - \log26\bigg) \:=\:2\!\cdot\!\log4 + \log26$

Therefore:. . . . . . . . . . . . $x \;=\;\dfrac{2\!\cdot\!\log4 + \log26}{3\!\cdot\!\log4 - \log26}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can be simplified beyond all recognition . . .

$x \;=\;\dfrac{\log(4^2) + \log(26)}{\log(4^3) - \log(26)}$

. . $=\;\dfrac{\log(16) + \log(26)}{\log(64) - \log(26)}$

. . $=\;\dfrac{\log(16\cdot26)}{\log\left(\frac{64}{26} \right)}$

. . $=\;\dfrac{\log(416)}{\log\left(\frac{32}{13}\right )}$

**Prove It** · July 21st 2010, 11:05 PM

Originally Posted by Soroban

Hello, porge111!

You have: . $3x\!\cdot\!\log4 - 2\!\cdot\!\log 4 \:=\:x\!\cdot\!\log26 + \log26$

. . . . . . . $3x\!\cdot\!\log4 - x\!\cdot\!\log26 \;=\;2\!\cdot\!\log4 + \log26$

Factor: . $x\!\cdot\!\bigg(3\log4 - \log26\bigg) \:=\:2\!\cdot\!\log4 + \log26$

Therefore:. . . . . . . . . . . . $x \;=\;\dfrac{2\!\cdot\!\log4 + \log26}{3\!\cdot\!\log4 - \log26}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can be simplified beyond all recognition . . .

$x \;=\;\dfrac{\log(4^2) + \log(26)}{\log(4^3) - \log(26)}$

. . $=\;\dfrac{\log(16) + \log(26)}{\log(64) - \log(26)}$

. . $=\;\dfrac{\log(16\cdot26)}{\log\left(\frac{64}{26} \right)}$

. . $=\;\dfrac{\log(416)}{\log\left(\frac{32}{13}\right )}$

Or as I put it...

$x = \frac{4\ln{(2)} + \ln{(26)}}{6\ln{(2)}- \ln{(26)}}$

which I prefer since the numbers inside the logarithms are the smallest they can possibly be.

**porge111** · July 22nd 2010, 08:45 AM

thanks guys!

I was just being silly as usual it seems.

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