# Exp. Equation

1. Jan 12, 2006

### cscott

$$e^{2x + 1} = 5$$

How can I solve this without a calculator?

2. Jan 12, 2006

### EnumaElish

Do you need a number or an expression as an answer?

3. Jan 12, 2006

### cscott

For the other parts of the question I've been able to find a numerical answer so I assume I should be finding one for this one, but is it possible without a calculator?

4. Jan 12, 2006

### Aneleh

Not unless you can do ln(5) in your head or somehow...:uhh: Usually an expression is enough, depends on how it's being marked though.

5. Jan 12, 2006

### Jameson

Well I get $$x=\frac{\ln(5)-1}{2}$$. That doesn't have an exact decimal representation, so you can leave it in exact form or approximate.

6. Jan 12, 2006

### cscott

Ah ok, thanks.

7. Jan 12, 2006

### benorin

The method of solution is as follows:

Given $$e^{2x + 1} = 5,$$

take the ln of both sides,

$$\ln \left( e^{2x + 1}\right) = \ln (5),$$

recall that $\ln \left( a^{x}\right) = x\ln \left( a\right)$, so we have

$$(2x + 1)\ln \left( e\right) = \ln (5),$$

and since $\ln \left( e\right) = 1$, we have

$$(2x + 1)(1) = \ln (5),$$

hence

$$x=\frac{1}{2}\left( \ln (5) -1\right)$$

8. Jan 12, 2006

### Hurkyl

Staff Emeritus
Sure it does!

Of course, I know you meant one that we can write with finitely many digits... but I don't want this to perpetuate the myth that decimal expansions do not exactly represent real numbers.