Natural Logs - Homework Question

I haven't taken a math class in over a year and a half and am in A Brief Survey of Calculus right now. Our current lesson is over Natural Logs, but I can't seem to figure out how to do this problem. If someone could please show me how to this it'd be amazing!

5*6^x = 4e^-5x

Re: Natural Logs - Homework Question

What have you tried so far?

Re: Natural Logs - Homework Question

I keep taking the natural log of both sides, which will bring both -5x and x down, then divide -5x by x and the x is cancelled out.

Re: Natural Logs - Homework Question

Ah. Your problem is that the constants on each side of the equation are *not* raised to a power. How does the logarithm handle multiplication? That is, you have

$\displaystyle \ln(5\cdot 6^{x})=\ln(4e^{-5x});$

what happens to the 5 on the LHS, and the 4 on the RHS?

Re: Natural Logs - Homework Question

Would you multiply the 5 into the 6? And the 4 into the e? What I end up getting looks like this before I'm lost.

x*ln7.5 = -5x

Re: Natural Logs - Homework Question

Quote:

Originally Posted by

**avonscooper** Would you multiply the 5 into the 6? And the 4 into the e? What I end up getting looks like this before I'm lost.

x*ln7.5 = -5x

No, I don't think you've go the idea yet. What is

$\displaystyle \ln(x\cdot y)?$

Re: Natural Logs - Homework Question

Hello, avonscooper!

Okay, here it is . . .

Quote:

$\displaystyle 5\cdot 6^x \:=\: 4e^{-5x}$

Take logs: .$\displaystyle \ln\left(5\cdot6^x\right) \:=\:\ln\left(4e^{-5x}\right) $

. . . . . .$\displaystyle \ln(5) + \ln\left(6^x\right) \:=\:\ln(4) + \ln\left(e^{-5x}\right)$

. . . . . .$\displaystyle \ln(5) + x\ln(6) \:=\:\ln(4) - 5x\underbrace{\ln(e)}$

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle ^{\text{This is 1}}$

. . . . . .$\displaystyle \ln(5) + x\ln(6) \:=\:\ln(4) - 5x$

. . . . . . . . $\displaystyle 5x + x\ln(6) \:=\:\ln(4) - \ln(5) \:=\:\ln\left(\tfrac{4}{5}\right) $

. . . . . . . . $\displaystyle x[5 + \ln(6)] \:=\:\ln(0.8)$

. . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{\ln(0.8)}{5 + \ln(6)} \:\approx\:-0.032855043 $

Re: Natural Logs - Homework Question

Quote:

Originally Posted by

**avonscooper** I haven't taken a math class in over a year and a half and am in A Brief Survey of Calculus right now. Our current lesson is over Natural Logs, but I can't seem to figure out how to do this problem. If someone could please show me how to this it'd be amazing!

5*6^x = 4e^-5x

You could alternatively try a few simplification steps,

to get down to a single "x".

$\displaystyle (5)6^x=4e^{-5x}=\frac{4}{e^{5x}}$

$\displaystyle 6^x=\frac{4}{5e^{5x}}$

$\displaystyle 6^xe^{5x}=\frac{4}{5}=0.8$

Now use the fact that

$\displaystyle (ab)^c=a^cb^c$

to get

$\displaystyle 6^x\left(e^5\right)^x=0.8$

$\displaystyle \left(6e^5\right)^x=0.8$

Now you can very simply take logs of both sides to solve for x.