Solving an Exponential Equation with Logarithm Rules

[jsully]

Having an issue with applying logarithm rules. I'm trying to find the value of an exponent in an equation of the form: A=B\frac{(1+x)^n-1}{x}
I have reviewed logarithm rules and cannot get to the answer, n, which is the only unkown variable in the equation. I've divided A by B, but am stuck at that point. Any assistance would be greatly appreciated.

Edit:

Just to be more clear I've plugged in some sample values in the event my variable choices were confusing:

230,000=1500\frac{(1+.00077)^n-1}{.00077}

I'm now at

153.33=\frac{(1+.00077)^n-1}{.00077}

...and stuck :( I think I need to log both sides, something like log 153.33=(n)log \frac{(1+.00077)-1}{.00077} .

 

[Square1]

If you divide by B, I think there is another step you could also come up with before the pace changes.

 

[Square1]

log153.33=(n)log(1+.00077)−1.00077

I think you are on the right track sort of but that step you took does not bring down the 'n'. Need to do some more to RHS before that can happen.

 

[jsully]

What do you mean by RHS?

 

[Square1]

right hand side

 

[jsully]

If I simplify the RHS further, isn't it just 1? \frac{(1+.00077)-1}{.00077}

 

[Square1]

A=B\frac{(1+x)^n-1}{x}

From there, to try to get the RHS to look like this:
(1+x)^n

To do that, you will have to simplify by adding, dividing, and multiplying values to both sides of the equation. Try and figure out the correct order and with what values that will work.

Once you are left with (1+x)^n , you can take logarithms. Then finish isolating n.

GTG though. GL.

 

[Mute]

If I simplify the RHS further, isn't it just 1? \frac{(1+.00077)-1}{.00077}

The problem that Square1 is trying to get you to notice is that you wrote something akin to $\ln(a^n + b) = n \ln(a +b)$, which is NOT correct. You need to isolate the "aⁿ" (in your case, the (1+x)ⁿ) before you apply the logarithm to both sides of your equation.

 

[Ratch]

jsully,

...I have reviewed logarithm rules and cannot get to the answer, n, which is the only unkown variable in the equation. ...

What are you trying to do? You have two variables in that expression, namely "x" and "n". If you specify x, then calculating n is trivial.

Ratch