Question

Y= a b^3x , where a>0 and b>0 solve for x ?

Answer 1

Assumption: you meant `y=ab^(3x)`

Hello Kevin.
If this is not correct please open in edit mode and adjust as necessary. Also look at how I have interpreted the solution options and correct as necessary.

Explanation:

You advise that the choice is:

"they gave me these to pick from" A, ln (y/a) / 3lnb. B, ln y/ 3ln a ln b.
C, ln y / 3ln(ab). D, 3lnb / ln y ln a E ,3ln b / ln (ay).

A:`color(white)("d") ln (y/a) / (3lnb)`

B`color(white)("d") ln y/ (3ln a ln b)`.

C`color(white)("d") ln y / (3ln(ab))`.

D`color(white)("d") (3lnb) / (ln y ln a)`

E`color(white)("d") (3ln b) / ln (ay)`

color(white)("d")
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Divide both side by `a`

`y/a=b^(3x)`

Take logs of both sides. I choose log to base `e ->ln`

`ln(y)-ln(a)=3xln(b)`

`x=(ln(y)-ln(a))/(3ln(b))`

However: `ln(y)-ln(a)` is the same as `ln(y/a)` giving:

`x=ln(y/a)/(2ln(b))->` solution A is the one you need

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Have a look at https://socratic.org/help/symbols

To obtain the format: `y=ab^(3x)color(white)("d")` I wrote:

hash symbol y=ab^(3x) hash symbol