# Thread: Solve x if b=(a^x+a^-x)/2

1. ## Solve x if b=(a^x+a^-x)/2

$\displaystyle \\b=\frac{a^x+a^{-x}}{2}\text{Find x}\\$
I took it to $\displaystyle x=\log_a (2b-a^{-x})$

The choices are
a)$\displaystyle log_a(b+\sqrt{b^2+1})$
b)$\displaystyle log_a(b+\sqrt{b^2-1})$
c)$\displaystyle log_b(a+\sqrt{a^2+1})$
d)none of these.

Is it none of these?
Can anyone help me? Or at least a hint?

2. ## Re: Solve x if b=(a^x+a^-x)/2

Originally Posted by NameIsHidden
$\\b=\frac{a^x+a^{-x}}{2}\text{Find x}\\$
I took it to $\displaystyle x=\log_a (2b-a^{-x})$

The choices are
a)$\displaystyle log_a(b+\sqrt{b^2+1})$
b)$\displaystyle log_a(b+\sqrt{b^2-1})$
c)$\displaystyle log_b(a+\sqrt{a^2+1})$
d)none of these.
Write as $a^{2x}-2ba^x+1=0$ solve for $a^x$.

BTW use $\$a^x\$$not [ tex][/tex]. 3. ## Re: Solve x if b=(a^x+a^-x)/2 Originally Posted by Plato Write as a^{2x}-2ba^x+1=0 solve for a^x. BTW use \a^x\$$ not [ tex][/tex].
$a^x=\frac{1}{a^x-2b}$

If I substitute this in given equation, I end up with weird results.

4. ## Re: Solve x if b=(a^x+a^-x)/2

Hello, NameIsHidden!

$\displaystyle b\:=\:\frac{a^x+a^{-x}}{2}.\;\text{ Find }x$

$\displaystyle (a)\;\log_a(b+\sqrt{b^2+1})$
$\displaystyle (b)\;\log_a(b+\sqrt{b^2-1})$
$\displaystyle (c)\;\log_b(a+\sqrt{a^2+1})$
$\displaystyle (d)\text{ none of these}$

I took it to $\displaystyle x\:=\:\log_a (2b-a^{-x})$
You haven't solve for x.

We have: .$\displaystyle b \;=\;\frac{a^x + a^{-x}}{2}$

Multiply by 2: .$\displaystyle 2b \;=\;a^x + a^{-x}$

Multiply by $\displaystyle a^x\!:\;2ba^x \;=\;a^{2x} + 1 \quad\Rightarrowquad a^{2x} - 2ba^x + 1 \;=\;0$

Quadratic Formula: .$\displaystyle a^x \;=\;\frac{2b \pm\sqrt{4b^2-4}}{2} \quad\Rightarrow\quad a^x \;=\;b \pm\sqrt{b^2-1}$

Take logs, base $\displaystyle a$: .$\displaystyle \log_a(a^x) \;=\;\log_a(b \pm \sqrt{b^2-1)$

Therefore: .$\displaystyle x \;=\;\log_a(b+ \sqrt{b-1})\qquad \text{answer (b)}$

5. ## Re: Solve x if b=(a^x+a^-x)/2

Originally Posted by Soroban
Hello, NameIsHidden!

We have: .$\displaystyle b \;=\;\frac{a^x + a^{-x}}{2}$

Multiply by 2: .$\displaystyle 2b \;=\;a^x + a^{-x}$

Multiply by $\displaystyle a^x\!:\;2ba^x \;=\;a^{2x} + 1 \quad\Rightarrowquad a^{2x} - 2ba^x + 1 \;=\;0$

Quadratic Formula: .$\displaystyle a^x \;=\;\frac{2b \pm\sqrt{4b^2-4}}{2} \quad\Rightarrow\quad a^x \;=\;b \pm\sqrt{b^2-1}$

Take logs, base $\displaystyle a$: .$\displaystyle \log_a(a^x) \;=\;\log_a(b \pm \sqrt{b^2-1)$

Therefore: .$\displaystyle x \;=\;\log_a(b+ \sqrt{b-1})\qquad \text{answer (b)}$
I wonder, do I need quadratic equation for this.

BTW thanks for your help.

6. ## Re: Solve x if b=(a^x+a^-x)/2

Originally Posted by NameIsHidden
I wonder, do I need quadratic equation for this.

BTW thanks for your help.
You never need the quadratic equation if you know other methods for solving quadratics. You can also try factoring or completing the square (completing the square is similar to using the quadratic equation, since the derivation of the quadratic equation involves completing the square of a general quadratic). However, in general, an equation $\displaystyle x = f(x)$ is called an implicit definition for $\displaystyle x$. When you write $\displaystyle x = \log_a(2b-a^{-x})$, you do not give an explicit definition for $\displaystyle x$. In order to have an explicit definition (to find the actual value for a variable), you need that variable on one side of the equation and the other side to be a function that does not include the variable you solved for. So, with Plato's and Soroban's solution, they have $\displaystyle x$ on one side of the equation and a function that does NOT involve $\displaystyle x$ on the other (making their solution an explicit definition for $\displaystyle x$, allowing them to find an actual value for it).

7. ## Re: Solve x if b=(a^x+a^-x)/2

You don't need to use the quadratic formula or even write it as a quadratic equation (if that was your question).

We can write $\displaystyle a^x= e^{ln(a^x)}= e^{xln(a)}$ and $\displaystyle a^{-x}= e^{ln(a^{-x})}= e^{-xln(a)}$

So $\displaystyle \frac{a^x+ a^{-x}}{2}= \frac{e^{xln(a)}+ e^{-xln(a)}}{2}= cosh(xln(a))= b$
so that $\displaystyle x= \frac{cosh^{-1}(b)}{ln(a)}$
Of course, that isn't one of the forms given but you can write $\displaystyle cosh^{-1}(b)$ in terms of a logarithm, basically by solving a quadratic equation as you have been told!