Finding compound interest rate

[coooool222]

To the nearest tenth, what interest rate, compounded annually, will produce $4613 if $3500 is left at this interest for 10 years?


Please don't give me the manipulated answer to answer the rate of interest. Heres what I did

4613 = 3500(1+R)^10

1.318= (1+R)^10

log 1.318 = 10 log (1+R)

Here is my work I'm stuck after this.

 

[JeffM]

I have not checked your arithmetic BUT

[math]\log_{\alpha}(\beta) = \log_{\alpha}(\gamma) \iff \beta = \gamma > 0.[/math]
Use that to finish up.

 

[coooool222]

I have not checked your arithmetic BUT

[math]\log_{\alpha}(\beta) = \log_{\alpha}(\gamma) \iff \beta = \gamma > 0.[/math]
Use that to finish up.

I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)

 

[BigBeachBanana]

I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)

I would divide by 10, then take the exponents on both sides to reverse the log.

 

[pka]

I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)

Because [imath]1.318=1318\cdot 10^{-3}[/imath] we get [imath]\log(1.318)=\log(1318)-3\log(10)[/imath]
[imath][/imath][imath][/imath][imath][/imath]

 

[JeffM]

I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)

Why bother with logs at all?

[math]1.318 = (1 + r)^{10} \implies \sqrt[10]{1.318} = 1 + r \implies r = \sqrt[10]{1.312} - 1.[/math]
EDIT: In the old days before calculators, we would of course have used logs to find the tenth root of 1.318, but that is 50 years anachronistic

 

[coooool222]

Why bother with logs at all?

[math]1.318 = (1 + r)^{10} \implies \sqrt[10]{1.318} = 1 + r \implies r = \sqrt[10]{1.312} - 1.[/math]

I was going to do that

 

[Steven G]

We have 1.318= (1+R)^10
If you want to use logs then log1.318= log(1+R)^10 (don't bring down the 10!)
(Since log A = Log B we get A=B)
log1.318= log(1+R)^10 so 1.318= (1+R)^10
Now solve this last equation for R.

Of course this begs the question why you introduced logs to begin with.