1

$$ \frac{1}{2^a} = 3.0988$$

How do I solve for $a$?

The original equation was:

$$\frac{1}{i^a}-\left(B-\frac{NE}{P_1-E}\right) = 0$$

I know that:

  • $B=10000$
  • $N = 50$
  • $E = 15$
  • $P_1 = 3000$

$$\frac1{i^a}-\left(10000-\frac{50(15)}{3000-15}\right) = 0$$

$$\frac1{i^a} - 3.0988 = 0$$

so if $i = 2$ how do I solve for $a$?

David G. Stork
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silversunhunter
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3 Answers3

1

$$2^{-a} = 3.0988$$

Taking logarithm,

$$-a\ln 2 = \ln 3.0988$$

$$a = -\frac{\ln 3.0988}{\ln 2}=-\log_2 3.0988$$

Siong Thye Goh
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1

Using $\frac{1}{2^a} = 3.0988$,

$$\begin{align} \ln\left(\frac {1}{2^a}\right) &= \ln\left(3.0988\right) \\ \ln\left(1\right) - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\ 0 - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\ \ln\left(2^a\right) &= - \ln\left(3.0988\right) \\ a\ln\left(2\right) &= -\ln\left(3.0988\right) \\ a &= \frac{- \ln\left(3.0988\right)} {\ln\left(2\right)} \end{align}$$

Or you simply work in $\log_2$:

$$\begin{align} \frac{1}{2^a} &= 3.0988 \\ 2^0 &= 2^a 2^{\log_2 3.0988 } \\ 0 &= a+ \log_2 3.0988 \\ a&=-\log_2 3.0988 \end{align}$$

gen-ℤ ready to perish
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user577215664
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    Math Machines answer was superb but the way you lay yours out helped me understand how to operate the solution with logk(n). Thank you! – silversunhunter Aug 31 '17 at 22:08
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    You're welcome....different poits of view help a lot and make thigs easier...next time you will know how to deal with these kind of math problems... – user577215664 Aug 31 '17 at 22:12
0

$\frac{1}{i^a}-\frac{10000-50(15)}{3000-15}=0$

$\frac{1}{i^a}-\frac{1850}{597}=0$

$\frac{1}{i^a}=\frac{1850}{597}$

i=2

$\frac{1}{2^a}=\frac{1850}{597}$

$2^a=\frac{597}{1850}$

I will now take the base two logarithm of both sides. If you already know what a logarithm is, skip 2 paragraphs down. In case you don't know, a logarithm is the opposite of a power. If you take the base two logarithm of a number, you're figuring out what power two has to be raised to to be that number.

$2^3=8$, so the base two logarithm of 8 is 3.

$3^5=243$, so the base three logarithm of 243 is 5.

To represent a base k logarithm of some number n, you write it as $log_k(n).$ most calculators dotn let you choose your base, and will only let you use a base 10 logarithm (represented log), or a base 2.7182818 logarithm (represented as ln). Luckily, though, you can find a base k logarithm by the simple equation, $log_k(n)=\frac{ln(n)}{ln(k)}$. Now, back to the equation.

$2^a=\frac{597}{1850}$

$log_2(2^a)=log_2(\frac{597}{1850})$

$a=log_2(\frac{597}{1850})$

In decimals, this can be represented as a=1.6317224341388696

Happy to help! If you want me to change any of the variables to make things less confusing or if you want me to explain anything you didn't get, just say so in the comments. Thanks!

Math Machine
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  • Hi. This is a great answer. I almost get it. So, since my calc does only ln i need to ln(2) / ln(597/1850) correct? – silversunhunter Aug 31 '17 at 22:01
  • No, it's $\frac{ln(\frac{597}{1850})}{ln(2)}$. You have to divide the logarithm of the number by the logarithm of the base. If you would like to see a demonstration as to why this works the way it does, I'd be glad to explain. – Math Machine Aug 31 '17 at 22:04
  • Ishams' answer helped me see a little clearer how to solve the equation. But your answer was robust and helped me understand more of the why.thank you! – silversunhunter Aug 31 '17 at 22:10