e4x+1= 2ex

2. ## Re: SOLVE FOR x please....

Originally Posted by Scoplex
e4x+1= 2ex
Can you solve $\displaystyle 3x+1=\log(2)~?$

3. ## Re: SOLVE FOR x please....

x = [log2 - 1]/3 ???

Does this mean that log(2) is equal to 2e?
I would then do the following:
4x+1=x
3x=-1
x=-1/3 ??? Is this then correct?

This logs and stuff get my head cooking sometimes.

4. ## Re: SOLVE FOR x please....

Originally Posted by Scoplex
x = [log2 - 1]/3 ???CORRECT

Does this mean that log(2) is equal to 2e?
What do you mean by that?

The $\displaystyle \log(2)$ is just $\displaystyle \log(2)$, no more, no less.

6. ## Re: SOLVE FOR x please....

Hello, Scoplex!

$\displaystyle \text{Solve for }x\!:\;\;e^{4x+1} \:=\: 2e^x$

Take logs: .$\displaystyle \ln\left(e^{4x+1}\right) \:=\:\ln(2e^x)$

. . . . . . $\displaystyle (4x+1)\ln(e) \:=\:\ln(2) + \ln\left(e^x\right)$

. . . . . . $\displaystyle (4x+1)\ln(e) \:=\:\ln(2) + x\ln(e)$ . . . Note that $\displaystyle \ln(e) = 1$

n . . . . . . . . . $\displaystyle 4x + 1 \:=\:\ln(2) + x$

n . . . . . . . . . $\displaystyle 3x + 1 \:=\:\ln(2)$

. . . . . . . . . . . . . $\displaystyle 3x \:=\:\ln(2) - 1$

. . . . . . . . . . . . . .$\displaystyle x \:=\:\dfrac{\ln(2) - 1}{3}$