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Hey guys could you please tell me what is the faster why to solve this equation. It's a compound interest equation and I'm stuck at the ${ (1 + 0.02) }^{ 24 }$ I really don't know how to proceed in this part. I thought about log, but how can I apply logarithm without touching on the $1500$?

Any help is appreciated, thanks.

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    $\begingroup$ Just divide both sides by $(1.02)^{24}$, no? $\endgroup$ Jun 10, 2013 at 12:34
  • $\begingroup$ @GerryMyerson there is any to solve ${ (1.02) }^{ 24 }$ without needing to multiply 1.02 with itself 24 times? $\endgroup$
    – Zignd
    Jun 10, 2013 at 12:45
  • $\begingroup$ in the calculation of compound interest it is often encounters.so this calculation is difficult to do manually.use calculator or you can do it by log table.If you want to simple this equation and use log table I can help then $\endgroup$ Jun 10, 2013 at 12:45
  • $\begingroup$ @iostream007 thanks for solving my doubt. $\endgroup$
    – Zignd
    Jun 10, 2013 at 12:47
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    $\begingroup$ Zignd, there is no way to "solve" $1.02^{24}$, because "solve" goes with "equation" or "problem", not with "number"; the word you are looking for is "evaluate", or "compute". As others note, evaluation can be done by repeated squaring, or using logarithms, or a calculator with the appropriate buttons; there are also websites where you can type in (1.02)^(24) and get an answer. $\endgroup$ Jun 11, 2013 at 8:59

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You have $P=\frac {1500}{1.02^{24}}$. To get $1.02^{24}$, you can use log tables to get $1.02^{24}=\exp(24 \ln 1.02)$. You can use repeated squaring to get$1.02^2, 1.04^4, \dots 1.02^{16}$ and finally use $1.02^{24}=1.02^{16}\cdot1.02^8$ which only takes five multiplies, or you can get a calculator that does powers. The final value is about $1.608$

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$1500=P \times { (1 + 0.02) }^{ 24 }$ if you don't want to touch $1500$ then just find out the value of ${ (1 + 0.02) }^{ 24 }$ .you can do it as in your selected answer and via log table like this: $$x={ (1 + 0.02) }^{ 24 }$$ take log on both side $$\log x=\log { (1 + 0.02) }^{ 24 }$$ $$\log x=24\log { (1.02) }$$ $$\log x=24\times 0.0086$$ $$\log x=0.2064$$ $$x=Antilog(0.2064)$$ antilog is a inverse function of $\log$ there is also an antilog table to see value.To calculate via calculator or manually it is like this: $$x=10^{0.2064}$$ $$x=1.6084$$ Now you have value of $(1.02)^{24}\;$which is $1.6084$ just put this value in equation : $$P=\dfrac{1500}{1.6084}\implies P=932.58$$

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To use a logarithm you would have to touch the 1500 [or 15,000, you reference both, we'll assume 1500]. It would look like

$$\log(1500)=24\times\log(P\times1.02)$$

Alternatively, if you wanted to approximate your solution you could use the binomial expansion with 0.02 as your small number and write something like

$$1500\approx P\times(1+24\times0.02)=P\times1.48$$

which is easily solved by hand. Otherwise using a calculator is your best option.

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  • $\begingroup$ The first order approximation is about $8.6\%$ low,which may be acceptable. It is a good thing to keep in mind. But OP can use logs just to handle the $1.02^{24}$, then divide into $1500$ $\endgroup$ Jun 10, 2013 at 13:15
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    $\begingroup$ Instead of a binomial expansion, you can get a more accurate approximation as $1.02^{24} = 1.02^{50/2} / 1.02 \approx \sqrt{e} / 1.02$. Since $e \approx 2.72$ we get $\sqrt{e} \approx 1.65$ and $1.65 / 1.02 \approx 1.62$, which is less than $0.7\%$ off. This is a bit more work though... $\endgroup$
    – TMM
    Jun 10, 2013 at 13:51

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