# Thread: Rational equation with X as an exponent, converting to a quadratic and solve w/log

1. ## Rational equation with X as an exponent, converting to a quadratic and solve w/log

I'll just start out by showing the equation.
$y = (10^x + 10^-^x) / (10^x - 10^-^x)$

I had a similar problem. I'll show how I worked that one out so that you can see my train of thought on this one.

$y = (10^x + 10^-^x)/2$
$2y = 10^x + 10^-^x$

Multiply both sides by $10^-^x$

$2y10^x = 10^2^x +1$

Now it's a quadratic. Solving for $10^x, a=1, b=-2y, and c=1$
$10^2^x - 2y^x +1 = 0$

I'll save both of us some time by not showing the quadratic equation steps.
So, I now have:
$10^x = y +- sqrt(y^2-1)$

Now, I take the log of both sides:
$log10^x = log(y +- sqrt(y^2-1))$

So, $x = log(y +- sqrt(y^2-1))$

Now, back to the problem at hand.

I multiplied both sides of the original equation by $10^x - 10^-^x$ to get:
$y10^x - y10^-^x = 10^x - 10^-^x$
I then multiplied both sides by $10^-^x$ to get:
$y10^2^x - y = 10^2^x - 1$

I then bring everything to one side:
$y10^2^x - 10^2^x - y + 1 = 0$

But, I can't see how this is quadratic. Either I screwed up before this point (the most likely scenario), or I'm just not seeing the quadratic.

2. Originally Posted by franzpulk
I then bring everything to one side:
$y10^2^x - 10^2^x - y + 1 = 0$
Now factor out 10^{2x}

$10^{2x}(y-1)- y + 1 = 0$

$10^{2x}(y-1)= y - 1$

$10^{2x}= \frac{y - 1}{y-1}$

$10^{2x}= 1 \implies x=0$

3. That seems a bit easy for problems from my instructor. But, it looks good to me. Thanks for the help.

4. Originally Posted by pickslides
Now factor out 10^{2x}

$10^{2x}(y-1)- y + 1 = 0$

$10^{2x}(y-1)= y - 1$

$10^{2x}= \frac{y - 1}{y-1}$

$10^{2x}= 1 \implies x=0$
0 isn't in the domain though? Since it would be division by 0?

If we sub in $x=0$ into $y = \dfrac{10^x + 10^-^x}{10^x - 10^{-x}}$ we get

$\dfrac{1+1}{1-1} = \dfrac{2}{0}$

5. x can equal zero in the domain as it's only an exponent. Anything raised to the power of zero is 1. 1 is allowed in the domain of the denominator, so x=0 is allowed.

6. Originally Posted by franzpulk
x can equal zero in the domain as it's only an exponent. Anything raised to the power of zero is 1. 1 is allowed in the domain of the denominator, so x=0 is allowed.
You're doing 1-1 in the denominator since exponents come before subtraction

Yours and Pickslides methods do work, but since 0 is not in the domain this equation has no solutions

edit: Your denominator is $10^x - 10^{-x}$. So for x =0 we get $10^0 - 10^{-0}$ and since $10^0 = 10^{-0} = 1$ then the denominator is $1-1 = 0$

7. Good point. So, there's something else to this problem then.

The example I showed on my first post was done in class. He said to make sure we paid attention to it. Then, he gives us homework, and while similar, it's nothing we've done before.

I'm still thinking that I need to make it a quadratic and solve it that way. Otherwise, why would he waste his time in class showing it to us?

Any tips? Also, don't rely on my math. Make sure you go by the original formula.

Thanks.

8. if the goal is to solve for x ...

$y(10^x - 10^{-x}) = 10^x + 10^{-x}
$

$y(10^{2x} - 1) = 10^{2x} + 1$

$10^{2x} \cdot y - y = 10^{2x} + 1$

$10^{2x} \cdot y - 10^{2x} = y + 1$

$10^{2x}(y - 1) = y+1$

$10^{2x} = \frac{y+1}{y-1}$

$2x = \log\left(\frac{y+1}{y-1}\right)$

$x = \log{\sqrt{\frac{y+1}{y-1}}}$

9. Originally Posted by franzpulk
I multiplied both sides of the original equation by $10^x - 10^{-x}$ to get:
$y10^x - y10^-^x = 10^x - 10^-^x$
I then multiplied both sides by $10^x$ to get:
$y10^2^x - y = 10^2^x - 1$
You seem to have had a sign change on the right hand side! When multiplying through by the denominator it looks like you changed your $+10^{-x}$ on the right to $-10^{-x}$

$y10^x - y10^{-x} = 10^x + 10^{-x}$

You carried forward the error but used the correct working (and some examiners will award you the method marks for this)

$y10^{2x} - y = 10^{2x} + 1$

$y10^{2x} - 10^{2x} = y + 1$

$10^{2x}(y-1) = y+1$

$10^{2x} = \dfrac{y+1}{y-1}$

$2x = \log \left(\dfrac{y+1}{y-1}\right)$

$x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)$

10. @ e^(i*pi) - That sign change was a typo on this post only. I have that correct on my hardcopy. I appreciate the help. What was getting me was pulling the $10^2^x$ out. That and not using the quadratic. Thanks again for the help.

@ skeeter - I'm assuming your last step was a typo? It should look like e^(i*pi)'s last step, correct? Not the square root like you showed? Thanks.

11. Originally Posted by franzpulk
@ skeeter - I'm assuming your last step was a typo? It should look like e^(i*pi)'s last step, correct? Not the square root like you showed?
no typo ... review your log properties.

12. @skeeter - I'm not finding anything regarding dividing a log and roots. You and e^(i*pi) have done two different things with the leading "2" coefficient:

$x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)$

and,

$x = \log{\sqrt{\frac{y+1}{y-1}}}$

I'm still not sure which is the correct method.

13. Originally Posted by franzpulk
I'll just start out by showing the equation.
$y = (10^x + 10^-^x) / (10^x - 10^-^x)$
Let $u= 10^x$
Then the equation becomes
$y= \frac{u+ \frac{1}{u}}{u- \frac{1}{u}}$
Multiplying numerator and denominator on the right by u
$y= \frac{u^2+ 1}{u^2- 1}$ so that $(u^2- 1)y= u^2y - u^2= u^2+ 1$

$(y- 2)u^2= 1$
$u^2= \frac{1}{y- 2}$
$u= 10^x= \frac{1}{\sqrt{y- 2}}$
$x ln(10)= -ln(\sqrt{y- 2})$

14. Originally Posted by franzpulk
@skeeter - I'm not finding anything regarding dividing a log and roots. You and e^(i*pi) have done two different things with the leading "2" coefficient:

$x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)$

and,

$x = \log{\sqrt{\frac{y+1}{y-1}}}$

I'm still not sure which is the correct method.
both are correct ...

the "power" property of logarithms states that if $y = a\log{x}$ , then $y = \log{x^a}$. The converse of this statement is also true.