Can someone tell me what I'm doing wrong (3((e^2x)/(e^x-3))+2=7)

[Ineedmathhelp123]

Problem: 3((e^2x)/(e^x-3))+2=7

What I did.

(3e^2x)/(3e^x-3)=5

3e^2x-(x-3)=5

e^2x-(x-3)=5/3

e^x-3=5/3

ln5/3=x-3

x=ln5/3+3

Answer from the book...
x=-3-ln3+ln5

I don't understand how to get there.....
Thanks for any help.

 

[mmm4444bot]

3e^(2x)/(e^x - 3) + 2 = 7

Note the grouping-symbol changes that I made above.

In the ratio, the numerator is 3 times a power of e, and the exponent is 2x.

The denominator is a difference: e^x minus 3.

Let us know, if this interpretation is not correct.

What I did.

3e^(2x)/(3e^x- 3) = 5

3e^(2x) - (x - 3) = 5

The factor of 3 is only in the numerator, but you multiplied both the numerator and the denominator by 3.

In the next step, the division by e^x-3 has changed to a subtraction of x-3. Is that what you meant to type?

 

[Ineedmathhelp123]

Note the grouping-symbol changes that I made above.

In the ratio, the numerator is 3 times a power of e, and the exponent is 2x.

The denominator is a difference: e^x minus 3.

Let us know, if this interpretation is not correct.


The factor of 3 is only in the numerator, but you multiplied both the numerator and the denominator by 3.

In the next step, the division by e^x-3 has changed to a subtraction of x-3. Is that what you meant to type?

Well this is what the problem looks like, Idk why but I couldn't upload a photo. Anyways middle problem.

https://prnt.sc/hnah38

Thanks for the help man.

 

[mmm4444bot]

\(\displaystyle 3 \bigg(\dfrac{e^{2x}}{e^{x - 3}}\bigg) + 2 = 7\)

Here is how we type it:

3e^(2x)/e^(x-3) + 2 = 7

My first two steps would be (1) subtract 2 from each side and (2) divide each side by 3.

Then you will have on the left-hand side a ratio of two powers of e. That is, the form:

e^n/e^m

This can be simplified, using the property of exponents which deals with ratios of powers of the same base. Do you remember the property?

Please show us how far you get. :cool:

 

[Ineedmathhelp123]

3e^(2x)/e^(x-3) + 2 = 7

e^2x-(x-3)=5/3 (quotient rule?)

e^2x-x+3=5/3

e^x+3=5/3

x+3=ln5/3

x=ln(5/3)-3

So I got to ln5/3-3=x

Is this an acceptable answer or do I have to turn it into -3-ln3+ln5=x, if I do how do I convert it.

 

[mmm4444bot]

3e^(2x)/e^(x-3) + 2 = 7

e^(2x-(x-3))=5/3 (quotient rule?)

e^(2x-x+3)=5/3

e^(x+3)=5/3

x+3=ln(5/3)

x=ln(5/3)-3

So I got to: ln(5/3) - 3 = x

Is this an acceptable answer …

Yes -- that's what I got.

If they want the other form, then they would need to say so (like: "Express the solution without any logarithms of fractions").

The property of logarithms used is:

ln(a/b) = ln(a) - ln(b)

You ought to also know ln(a·b) = ln(a) + ln(b) :cool: