Please help me solve these 2 logarithmic equations:
Thanks
$\ln\bigg[x^{3\ln{x}-2}\bigg] = \ln\bigg[\dfrac{x^{1+\ln{x}}}{e}\bigg]$
$(3\ln{x}-2)\ln{x} = (1+\ln{x})\ln{x} - 1$
$3\ln^2{x} - 2\ln{x} = \ln{x} + \ln^2{x} - 1$
$2\ln^2{x} - 3\ln{x} + 1 = 0$
$(2\ln{x} - 1)(\ln{x} - 1) = 0$
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.
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.
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?
$\text {Let } u = log_2(x).$
$x^{\frac{log_2(x) - 6}{4}} = \dfrac{4}{x} \implies log_2 \left ( x^{\frac{log_2(x) - 6}{4}} \right ) = log_2 \left ( \dfrac{4}{x} \right ) \implies$
$ \dfrac{log_2(x) - 6}{4} * log_2(x) = log_2(4) - log_2(x) \implies \dfrac{u - 6}{4} * u = 2 - u \implies$
$u^2 - 6u = 8 - 4u \implies what?$
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.