natural logarithm: express y = 1.6(0.2)^x in terms of e

[varga]

Rewrite the following equation in terms of base e. Express the answer in terms of a natural logarithm and then round to 3 decimal places.

y=1.6(0.2)* *---to the x power

Type the model in base of e

y= ? (round numbers to 3 decimal places)

 

[stapel]

x= ?

Why are you trying to solve the equation for x in terms of y? Aren't you supposed to be restating the equation in terms of e...? :shock:

Rewrite the following equation in terms of base e. Express the answer in terms of a natural logarithm and then round to 3 decimal places.

y=1.6(0.2)---to the x power

You have:

. . . . .y = 1.6(0.2)^x

Taking the log of each side, you get:

. . . . .ln(y) = ln(1.6) + xln(0.2)

Then raise both sides as powers on e, and you have the restated form of the equation. :wink:

Eliz.

 

[varga]

still confused...

still lost

 

[Deleted member 4993]

Let's do another problem then:

\(\displaystyle y = A\;\cdot\ B^x\)[ where 'A' & 'B' are some constants]

\(\displaystyle ln(y)\) = \(\displaystyle ln (A) + ln(B^x)\)

\(\displaystyle ln(y)\) = \(\displaystyle ln (A) + x\cdot\ln(B)\)

\(\displaystyle e^{ln(y)}\) = \(\displaystyle e^{[ln (A) + x\cdot\ln(B)]}\)

\(\displaystyle e^{ln(y)}\) = \(\displaystyle e^{ln (A)} \cdot\ e^{[ x\cdot\ln(B)]}\)

Now finish it by putting numbers given to you (for A & B)

 

[skeeter]

1.6 = e^ln(1.6)

0.2^x = [e^ln(0.2)]^x

so ... 1.6(0.2)^x = ?

 

[jwpaine]

Re: still confused...

still lost

By now you have been provided with three excellent examples that you can use to apply the concept to your own problem.

If you cannot understand this then I would recommend sitting down after class with your teacher.

Happy logging!
John