Solving with logarithms Solve $y = \left(\frac{10^x + 10^{-x}}{2}\right)$ for $x$ in terms of $y$.
I tried taking the log of both sides and got
$$\log(y) = \log(10^x + 10^{-x})  - \log(2)$$
Now I don't know what to do, please help?
 A: Let $a=(10)^x$.
the equation becomes
$$a+\frac{1}{a}=2y \;\;(>0)$$
or
$$a^2-2ay+1=0$$
the reduced discriminant is
$$\delta=y^2-1$$
so, there are three cases :


*

*$0<y<1$  there is no solution.

*$y=1 \implies a=1\implies x=0.$

*$y>1\implies a=y\pm\sqrt{y^2-1}$
$$\implies x=\frac{\ln(y\pm\sqrt{y^2-1})}{\ln(10)}.$$
Observe that if $x$ is a solution, $-x$ is also a solution.
A: We have:
$$
y=\frac{1}{2} \left(10^{-x}+10^x\right)
$$
First, we recognize that this is a hyperbolic cosine function:
$$
y=\cosh (x \ln (10))
$$
There is the inverse hyperbolic cosine function:
$$
x \ln (10)=\cosh^{-1} (y)
$$
Expressing in terms of a logarithm:
$$
x \ln (10)=\ln \left(y+\sqrt{y-1} \sqrt{y+1}\right)
$$
We can change the base of the logarithm:
$$
x =\log \left(y+\sqrt{y-1} \sqrt{y+1}\right)
$$
A: HINT:
Let $u=10^x$ and $u^{-1}=10^{-x}$.  Solve a quadratic equation for $u$, the take the logarithm of $u$ and solve for $x$.
A: With logarithms:
Let $x=\log_{10}t$. The equation becomes
$$t+\frac1t=2y.$$
Solving for $t$ (via a quadratic equation), you get
$$t=y\pm\sqrt{y^2-1}$$
and 
$$x=\log_{10}\left(y\pm\sqrt{y^2-1}\right).$$
The solutions exsist whenever $y\ge1$.
