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**yuriythebest** · May 19th 2008, 11:06 AM

sorry that I couldn't make the title more descriptive.

anyhow I've almost solved an equation and I now have

what's next? how do I get x out of this?

also

if I have 10^x - 5*4^x + 10=0

what can I possibly do next?

**Moo** · May 19th 2008, 11:45 AM

Hello

Originally Posted by yuriythebest

sorry that I couldn't make the title more descriptive.

anyhow I've almost solved an equation and I now have

what's next? how do I get x out of this?

Taking the logarithms :
$- \frac 1x \ln 8=\ln \frac{25}{2}=\ln 25 - \ln 2$

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

)

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

The equation is now :

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

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

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

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

**colby2152** · May 19th 2008, 11:46 AM

Originally Posted by yuriythebest

sorry that I couldn't make the title more descriptive.

anyhow I've almost solved an equation and I now have

what's next? how do I get x out of this?

also

if I have 10^x - 5*4^x + 10=0

what can I possibly do next?

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

Take everything to the "x" power.

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

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

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

Take the natural log of both sides

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

Can you solve that?

**Soroban** · May 19th 2008, 01:03 PM

Hello, yuriythebest!

Another approach . . .

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

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

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

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

**Soroban** · May 19th 2008, 01:27 PM

Heloo again, yuriythebest!

The second problem is rather bizarre.
I will assume we want integer solutions.

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

I found no direct approach to it . . .

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

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

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

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

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

**Reckoner** · May 19th 2008, 03:17 PM

As usual, Soroban, your post is very useful. However, the left side of your equation is never actually 1. The highlighted element should be $10^{1 - 1} + 1 = 10^0 + 1 = 2$ .

So, we really have

$ \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 $x\approx1.364841930549761$ , determined by Newton's method.

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