Thanks

2. ## Re: Logarithmic Equation

$\ln\bigg[x^{3\ln{x}-2}\bigg] = \ln\bigg[\dfrac{x^{1+\ln{x}}}{e}\bigg]$

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

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

$2\ln^2{x} - 3\ln{x} + 1 = 0$

$(2\ln{x} - 1)(\ln{x} - 1) = 0$

can you finish?

for the second, taking the base 2 log of each side yields ...

$\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

you'll get a quadratic in $\log_2{x}$ ...

give it a go.

3. ## Re: Logarithmic Equation

Originally Posted by IloveIl
1) OR rewrite as
\begin{align*}e\cdot{\left( x \right)^{3\log (x) - 2}} &= {x^{1 + \log (x)}} \\ e\cdot{\left( x \right)^{2\log (x) - 3}} &= 1\\1 + \left( {{2{\log }^2}(x) - 3\log (x)} \right) &= 0\end{align*}

Now let $y=\log(x)$ we get $2y^2-3y+1=0$ to solve.

Check Here.

4. ## Re: Logarithmic Equation

for the second, taking the base 2 log of each side yields ...

$\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

you'll get a quadratic in $\log_2{x}$ ...

give it a go.[/QUOTE]

I didn't understand at all what you did here. can you show in more details?

5. ## Re: Logarithmic Equation

Originally Posted by IloveIl
for the second, taking the base 2 log of each side yields ...

$\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

you'll get a quadratic in $\log_2{x}$ ...

give it a go.

I didn't understand at all what you did here. can you show in more details?
$\text {Let } u = log_2(x).$

$x^{\frac{log_2(x) - 6}{4}} = \dfrac{4}{x} \implies log_2 \left ( x^{\frac{log_2(x) - 6}{4}} \right ) = log_2 \left ( \dfrac{4}{x} \right ) \implies$

$\dfrac{log_2(x) - 6}{4} * log_2(x) = log_2(4) - log_2(x) \implies \dfrac{u - 6}{4} * u = 2 - u \implies$

$u^2 - 6u = 8 - 4u \implies what?$

6. ## Re: Logarithmic Equation

Originally Posted by IloveIl
for the second, taking the base 2 log of each side yields ...

$\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

you'll get a quadratic in $\log_2{x}$ ...

give it a go.
https://www.freemathhelp.com/forum/t..._2-x-6)-4)-4-x

IloveIl, you never responded back at this site with any attempt to show that you understood those helpers.

Had you followed up on that, then you might not be here asking the same questions.