# Thread: logs and exponents of natural base

1. ## logs and exponents of natural base

I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

1. Solve each equation
(a) e^x - 2e^-x = 1

(b) ln(x^2 + e) - ln (x + 1) = 1

If you could explain the steps in getting the answers, that would be much appreciated. I'm a little lost on how to start them off.

The answers we were given on the sheet are:
(a) ln2 (b) 0, e

2. Originally Posted by jmailloux
I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

1. Solve each equation
(a) e^x - 2e^-x = 1

(b) ln(x^2 + e) - ln (x + 1) = 1

If you could explain the steps in getting the answers, that would be much appreciated. I'm a little lost on how to start them off.

The answers we were given on the sheet are:
(a) ln2 (b) 0, e
a) e^x -2e^(-x) = 1
Rewriting that in fraction form,
e^x -2/(e^x) = 1
Clear the fraction, multiply both sides by e^x,
(e^x)^2 -2 = e^x
(e^x)^2 -e^x -2 = 0
(e^x -2)(e^x +1) = 0
e^x = 2 or -1

When e^x = 2,
x*ln(e) = ln(2)
x = ln(2) ---------------------**

When e^x = -1
x*ln(e) = ln(-1)
Uh-oh, there is no log of negative numbers, so disregard e^x = -1

---------------------------------------------------
b) ln(x^2 +e) -ln(x +1) = 1
ln[(x^2 +e) /(x +1)] = ln(e)
(x^2 +e) /(x +1) = e
x^2 +e = e(x+1)
x^2 +e = ex +e
x^2 = ex
x^2 -ex = 0
x(x -e) = 0
x = 0 or e. ----------------answer.

3. Originally Posted by jmailloux
I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

1. Solve each equation
(a) e^x - 2e^-x = 1
Multiply through by $\displaystyle e^x$ to get:

$\displaystyle (e^x)^2-2=e^x$

which is a quadratic in $\displaystyle y=e^x$ which making this substitution and re arranging gives:

$\displaystyle y^2-y-2=0$

which can be solved using the quadratic formula or by noticing that it factors to give:

$\displaystyle y^2-y-2=(y+1)(y-2)=0$

which has roots $\displaystyle y=-1,2$, s the solutions of the original equation are $\displaystyle x=\ln(-1)$ which is imposible (in the reals) and $\displaystyle x=\ln(2)$.

So the only solution in the real numbers is $\displaystyle x=\ln(2)$.

(b) ln(x^2 + e) - ln (x + 1) = 1
Here we use the law of logarithms to combine the terms on the left:

$\displaystyle \ln(x^2 + e) - \ln (x + 1)=\ln \left( \frac{x^2+e}{x+1} \right) = 1$

Now we use the definition of natural logarithms to get rid of the logs:

$\displaystyle \frac{x^2+e}{x+1}=e^1=e$

which is a quadratic in x:

$\displaystyle x^2+e=e\,x+e$

rearranging:

$\displaystyle x^2-e\,x=0$,

so $\displaystyle x=0, e$.

RonL

4. ## thanks

thanks.. it makes A LOT more sense now