1. ## logarithmic equation

${27x}^{\log_5 x}={x}^{4}$

2. ## Re: logarithmic equation

Originally Posted by Chipset3600
${27x}^{\log_5 x}={x}^{4}$
Let

$y=\log_{5}(x) \iff x=5^y$ Then you equation becomes

$27(5^y)^y=(5^y)^4$

$27 \cdot 5^y^2=5^{4y} \iff 27 \cdot 5^{y^2-4y}=1$

Now if we take log base 5 of both sides we get

$y^2-4y+\log_{5}(27)=0$

Solving this quadratic for y gives

$y=\frac{4 \pm \sqrt{16-4\log_{5}(27)}}{2}=2 \pm \sqrt{4-\log_{5}(27)}$

Pluging this back in we get the solution for x is

$x=5^{2 \pm \sqrt{4-\log_{5}(27)}}$

3. ## Re: logarithmic equation

Originally Posted by Chipset3600
${27x}^{\log_5 x}={x}^{4}$
Here's an alternative method...

\displaystyle \begin{align*} 27x^{\log_5{x}} &= x^4 \\ \log_x{\left( 27x^{\log_5{x}} \right)} &= \log_x{\left(x^4\right)} \\ \log_x{(27)} + \log_x{\left( x^{\log_5{x}}\right)} &= 4 \\ \log_x{(27)} + \log_5{x} &= 4 \\ \frac{\log_5{(27)}}{\log_5{x}} + \log_5{x} &= 4 \\ \log_5{(27)} + \log_5^2{x} &= 4\log_5{x} \end{align*}

4. ## Re: logarithmic equation

Hello, Chipset3600!

And yet another method . . .

$\displaystyle 27x^{\log_5 x}\:=\:x^4$

Take logs, base 5:

. . . . . . . . . . . $\log_5\left(27x^{\log_5x}\right) \;=\;\log_5(x^4)$

. . . . . . $\log_527 + \log_5\left(x^{\log_5x}\right) \;=\;4\log_5x$

. . . . . . $\log_527 + \log_5x\!\cdot\!\log_5x \;=\;4\log_5x$

. . $\left(\log_5x\right)^2 - 4\log_5x + \log_527 \;=\;0$

which is the quadratic equation found by Prove It.

5. ## Re: logarithmic equation

I understood all steps, but how can i solve that $\sqrt{4-\log_{5}(27)}$ without calculator?

6. ## Re: logarithmic equation

$3^{3}x^{\log_3 x } = x^{4}$

$x^{\log_3 x } = \frac{x^{4}}{27} doing \log_3 x = y so 3^{y} = x$

$27(3^{y})^{y} = (3^{y})^{y}$

$3^{3}.3^{y^{2}}= 3^{4y}$
$3^{3+y^{2}}= 3^{4y}$
$3+y^{2}=3^{4y}$
$3+y^{2}=4y$
$y^{2}-4y+3=0$
$y_{1}=1$ $or$ $y_{2}=3 \log_3 x =1 and \log_3 x = 3$

So x = 3 or x=27

7. ## Re: logarithmic equation

Sorry guys! i did a mistake, the base of the Log is 3 and not 5...Thanks