Natural Logs - Homework Question

**avonscooper** · June 25th 2011, 05:01 PM

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

**Ackbeet** · June 25th 2011, 05:22 PM

What have you tried so far?

**avonscooper** · June 25th 2011, 05:38 PM

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.

**Ackbeet** · June 25th 2011, 05:49 PM

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

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

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

**avonscooper** · June 25th 2011, 05:52 PM

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

**Ackbeet** · June 25th 2011, 06:15 PM

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

$\ln(x\cdot y)?$

**Soroban** · June 25th 2011, 07:03 PM

Hello, avonscooper!

Okay, here it is . . .

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

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

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

. . . . . . $\ln(5) + x\ln(6) \:=\:\ln(4) - 5x\underbrace{\ln(e)}$
. . . . . . . . . . . . . . . . . . . . . . . . . $^{\text{This is 1}}$

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

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

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

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

**Archie Meade** · June 26th 2011, 06:48 AM

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".

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

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

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

Now use the fact that

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

to get

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

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

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

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