# Logarithmic Equation

• May 3rd 2010, 05:15 AM
alternative
Logarithmic Equation
Hi, Could anyone help me solve this...?

log3 + log3 = 0
4x+1 3x

Thanks:D:D
• May 3rd 2010, 05:29 AM
Quote:

Originally Posted by alternative
Hi, Could anyone help me solve this...?

log3 + log3 = 0
4x+1 3x

Thanks:D:D

hi

i assume its $\displaystyle \log_3(4x+1)+\log_3(3x)=0$

If you are not sure how to generate these codes , click on it to see the codes .

Using the properties of log , where log a + log b = log (ab)

$\displaystyle \log_3(4x+1)(3x)=0$

Then convert this to indices form ,

$\displaystyle (4x+1)(3x)=3^0=1$

Solve for x then .
• May 3rd 2010, 09:31 AM
Soroban
Hello, alternative!

Is this what you meant?

Quote:

$\displaystyle \log_{4x+1}(3) + \log_{3x}(3) \;=\;0$

Use the Base-Change Formula: . $\displaystyle \frac{\ln(3)}{\ln(4x+1)} + \frac{\ln(3)}{\ln(3x)} \;=\;0$

Multiply by $\displaystyle \frac{\ln(4x+1)\ln(3x)}{\ln(3)}\!:\quad \ln(3x) + \ln(4x+1) \;=\;0 \quad\Rightarrow\quad \ln(4x+1) \:=\:-\ln(3x)$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle \ln(4x+1) \;=\;\ln(3x)^{-1} \quad\Rightarrow\quad \ln(4x+1) \;=\;\ln\left(\frac{1}{3x}\right)$

Expnentiate both sides: .$\displaystyle 4x + 1 \:=\:\frac{1}{3x} \quad\Rightarrow\quad 12x^2 + 3x - 1 \:=\:0$

Quadratic Formula: .$\displaystyle x \:=\:\frac{-3 \pm\sqrt{57}}{24}$

The negative root, $\displaystyle \frac{-3 - \sqrt{57}}{24}$, is extraneous.

The base of a logarithm must be positive.

Therefore: .$\displaystyle x \;=\;\frac{-3 + \sqrt{57}}{24}$

• May 3rd 2010, 12:38 PM
alternative
Quote:

Originally Posted by Soroban
Hello, alternative!

Is this what you meant?

Use the Base-Change Formula: . $\displaystyle \frac{\ln(3)}{\ln(4x+1)} + \frac{\ln(3)}{\ln(3x)} \;=\;0$

Multiply by $\displaystyle \frac{\ln(4x+1)\ln(3x)}{\ln(3)}\!:\quad \ln(3x) + \ln(4x+1) \;=\;0 \quad\Rightarrow\quad \ln(4x+1) \:=\:-\ln(3x)$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle \ln(4x+1) \;=\;\ln(3x)^{-1} \quad\Rightarrow\quad \ln(4x+1) \;=\;\ln\left(\frac{1}{3x}\right)$

Expnentiate both sides: .$\displaystyle 4x + 1 \:=\:\frac{1}{3x} \quad\Rightarrow\quad 12x^2 + 3x - 1 \:=\:0$

Quadratic Formula: .$\displaystyle x \:=\:\frac{-3 \pm\sqrt{57}}{24}$

The negative root, $\displaystyle \frac{-3 - \sqrt{57}}{24}$, is extraneous.

The base of a logarithm must be positive.

Therefore: .$\displaystyle x \;=\;\frac{-3 + \sqrt{57}}{24}$

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