# 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.
$\displaystyle y = (10^x + 10^-^x) / (10^x - 10^-^x)$

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.

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

Multiply both sides by $\displaystyle 10^-^x$

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

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

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

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

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

Now, back to the problem at hand.

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

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

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:
$\displaystyle y10^2^x - 10^2^x - y + 1 = 0$
Now factor out 10^{2x}

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

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

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

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

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}

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

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

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

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

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

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

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

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 ...

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

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

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

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

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

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

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

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

9. Originally Posted by franzpulk
I multiplied both sides of the original equation by $\displaystyle 10^x - 10^{-x}$ to get:
$\displaystyle y10^x - y10^-^x = 10^x - 10^-^x$
I then multiplied both sides by $\displaystyle 10^x$ to get:
$\displaystyle y10^2^x - y = 10^2^x - 1$
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 $\displaystyle +10^{-x}$ on the right to $\displaystyle -10^{-x}$

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

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

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

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

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

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

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

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

Still no quadratic though

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 $\displaystyle 10^2^x$ 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:

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

and,

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

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

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

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

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:

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

and,

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

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

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