# logarithmic equation

• Oct 3rd 2012, 06:33 PM
Chipset3600
logarithmic equation
$\displaystyle {27x}^{\log_5 x}={x}^{4}$
• Oct 3rd 2012, 07:56 PM
TheEmptySet
Re: logarithmic equation
Quote:

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

Let

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

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

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

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

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

Solving this quadratic for y gives

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

Pluging this back in we get the solution for x is

$\displaystyle x=5^{2 \pm \sqrt{4-\log_{5}(27)}}$
• Oct 3rd 2012, 08:36 PM
Prove It
Re: logarithmic equation
Quote:

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

Here's an alternative method...

\displaystyle \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*}

which is now a quadratic.
• Oct 3rd 2012, 09:54 PM
Soroban
Re: logarithmic equation
Hello, Chipset3600!

And yet another method . . .

Quote:

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

Take logs, base 5:

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

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

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

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

which is the quadratic equation found by Prove It.
• Oct 4th 2012, 02:19 AM
Chipset3600
Re: logarithmic equation
I understood all steps, but how can i solve that $\displaystyle \sqrt{4-\log_{5}(27)}$ without calculator?
• Oct 4th 2012, 05:50 AM
Chipset3600
Re: logarithmic equation
$\displaystyle 3^{3}x^{\log_3 x } = x^{4}$

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

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

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

So x = 3 or x=27
• Oct 4th 2012, 01:40 PM
Chipset3600
Re: logarithmic equation
Sorry guys! i did a mistake, the base of the Log is 3 and not 5...Thanks