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- May 19th 2008, 10:06 AM #1

- May 19th 2008, 10:45 AM #2
Hello

Taking the logarithms :

$\displaystyle - \frac 1x \ln 8=\ln \frac{25}{2}=\ln 25 - \ln 2$

But $\displaystyle 8=2^3 \implies \ln 8=3 \ln 2$ (it's more beautiful )

And $\displaystyle 25=5^2 \implies \ln 25=2 \ln 5$

The equation is now :

$\displaystyle -\frac 1x (3\ln 2)=2 \ln 5-\ln 2$

$\displaystyle - \frac 1x=2 \frac{\ln 5}{3\ln 2}-\frac 13$

$\displaystyle \frac 1x=\frac 13-2\frac{\ln 5}{3 \ln 2}$

$\displaystyle x=\frac{1}{\frac 13-2\frac{\ln 5}{3 \ln 2}} \approx -0.82330362225011$

- May 19th 2008, 10:46 AM #3
$\displaystyle 8^{\frac{-1}{x}}=\frac{25}{2}$

Take everything to the "x" power.

$\displaystyle 8^{\frac{-x}{x}}=\left(\frac{25}{2}\right)^x$

$\displaystyle 8^{-1}=\left(\frac{25}{2}\right)^x$

$\displaystyle \frac{1}{8}=\left(\frac{25}{2}\right)^x$

Take the natural log of both sides

$\displaystyle \ln \left(\frac{1}{8}\right)=x \ln \left(\frac{25}{2}\right)$

Can you solve that?

- May 19th 2008, 12:03 PM #4

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Hello, yuriythebest!

Another approach . . .

$\displaystyle 8^{-\frac{1}{x}} \:=\:\frac{25}{2}$

Take reciprocals: .$\displaystyle 8^{\frac{1}{x}} \:=\:\frac{2}{25} \:=\:\frac{8}{100}$

Take logs (base 10): .$\displaystyle \log\left(8^{\frac{1}{x}}\right) \:=\:\log\left(\frac{8}{100}\right)\quad\Rightarro w\quad \frac{1}{x}\log(8) \:=\:\log(8) - 2$

Hence: .$\displaystyle \frac{1}{x} \:=\:\frac{\log(8) - 2}{\log(8)} \quad\Rightarrow\quad\boxed{ x \;=\;\frac{\log(8)}{\log(8) - 2}} $

- May 19th 2008, 12:27 PM #5

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Heloo again, yuriythebest!

The second problem is rather bizarre.

I will assume we want*integer*solutions.

$\displaystyle 10^x - 5*4^x + 10\:=\:0$

We have: .$\displaystyle 10^x + 10 \:=\:5\cdot4^x \quad\Rightarrow\quad 10\left(10^{x-1}+1\right) \:=\:5\cdot2^{2x}$

Divide by 10: .$\displaystyle 10^{x-1}+1 \;=\;2^{2x-1}$

The left side is of the form: .$\displaystyle \hdots\:1.001,\:1.01,\:1.1,\:{\color{red}1},\:11,\ :101,\:1001,\:\hdots$

The right side is of the form: .$\displaystyle \hdots\:\frac{1}{8},\:\frac{1}{4},\:\frac{1}{2},\: {\color{red}1},\:2,\:4,\:8,\:\hdots$

They are equal only when: .$\displaystyle \begin{array}{ccc}10^{x-1} + 1 &=& 1 \\ 2^{2x-1} &=& 1\end{array}\quad\Rightarrow\quad\boxed{x \:=\:1}$

- May 19th 2008, 02:17 PM #6
As usual, Soroban, your post is very useful. However, the left side of your equation is never actually 1. The highlighted element should be $\displaystyle 10^{1 - 1} + 1 = 10^0 + 1 = 2$.

So, we really have

$\displaystyle

\begin{array}{ccc}

10^{x-1} + 1 &=& 2\\

2^{2x-1} &=& 2

\end{array}\quad\Rightarrow\quad\boxed{x \:=\:1}

$

Of course, you still had the right solution.

Incidentally, the other solution should be about $\displaystyle x\approx1.364841930549761$, determined by Newton's method.