Can You Solve e^(2x+1) = 5 Without a Calculator?

[cscott]

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

How can I solve this without a calculator?

 

[EnumaElish]

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

 

[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?

 

[Aneleh]

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

 

[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.

 

[cscott]

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.

Ah ok, thanks.

 

[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

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

 

[Hurkyl]

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.

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.