1. ## Logarithmic Equation

Hi, Could anyone help me solve this...?

log3 + log3 = 0
4x+1 3x

Thanks

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

log3 + log3 = 0
4x+1 3x

Thanks
hi

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

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

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

Then convert this to indices form ,

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

Solve for x then .

3. Hello, alternative!

Is this what you meant?

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

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

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

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

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

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

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

The base of a logarithm must be positive.

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

4. Originally Posted by Soroban
Hello, alternative!

Is this what you meant?

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

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

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

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

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

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

The base of a logarithm must be positive.

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

I find no words to thank you......