Finding compound interest rate

coooool222

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Jun 1, 2020
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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.
 
I have not checked your arithmetic BUT

logα(β)=logα(γ)    β=γ>0.\log_{\alpha}(\beta) = \log_{\alpha}(\gamma) \iff \beta = \gamma > 0.
Use that to finish up.
 
I have not checked your arithmetic BUT

logα(β)=logα(γ)    β=γ>0.\log_{\alpha}(\beta) = \log_{\alpha}(\gamma) \iff \beta = \gamma > 0.
Use that to finish up.
I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)
 
I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)
Because 1.318=13181031.318=1318\cdot 10^{-3} we get log(1.318)=log(1318)3log(10)\log(1.318)=\log(1318)-3\log(10)
 
I don't understand. I am stuck in solving this log 1.318 = 10 log (1+R)
Why bother with logs at all?

1.318=(1+r)10    1.31810=1+r    r=1.312101.1.318 = (1 + r)^{10} \implies \sqrt[10]{1.318} = 1 + r \implies r = \sqrt[10]{1.312} - 1.
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
 
Why bother with logs at all?

1.318=(1+r)10    1.31810=1+r    r=1.312101.1.318 = (1 + r)^{10} \implies \sqrt[10]{1.318} = 1 + r \implies r = \sqrt[10]{1.312} - 1.
I was going to do that :cautious:
 
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.
 
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