1. ## Solve Logarithmic Equations

Needing help with a homework question. I know the answer but do not know how to show my work to get that answer:
Ln(4x^2-3x)=ln(16x-12)-ln x

2. ## Re: Solve Logarithmic Equations

$Ln(4x^2-3x)=ln(16x-12)-ln x$ This can be written as : $Ln(4x^2-3x)=ln ( \frac{(16x-12)}{x})${ using property of log a - log b = log (a/b) } Now we can compare both sides we get : $4x^2-3x = \frac{(16x-12)}{x}$ Now you can easily solve this equation. Hope you understood

3. ## Re: Solve Logarithmic Equations

Hello, paigesisco!

$\text{Solve: }\:\ln(4x^2-3x)\:=\:\ln(16x-12)-\ln x$

Note that: . $16x-12\:>\:0 \quad\Rightarrow\quad x \:>\:\tfrac{3}{4}$

We have: . $\ln(4x^2-3x) \ln x \:=\:\ln(16x-12)$

. . . . . . . . $\ln\left[x(4x^2 - 3x)\right] \:=\:\ln(16x-12)$

. . . . . . . . . . $\ln(4x^3 - 3x^2) \:=\:\ln(16x-12)$

. . . . . . . . . . . . $4x^3 - 3x^2 \:=\:16x-12$

. . . . $4x^3 - 3x^2 - 16x + 12 \:=\:0$

. . . $x^2(4x-3) - 4(4x - 3) \:=\:0$

. . . . . . . $(x^2-4))(4x-3) \:=\:0$

Hence: . $x \;=\;\pm2,\,\tfrac{3}{4}$

The only root in the domain is: . $x = 2$