Solving with logarithms

Question

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?

Answer 1

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.

Answer 2

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 $x$.

Answer 3

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: $$ x =\log \left(y+\sqrt{y-1} \sqrt{y+1}\right) $$

Answer 4

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 $y\ge1$.