How do I find an unknown power in this formula?

Question

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

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 $a$?

Answer 1

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

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

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

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!

Answer 2

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

Taking logarithm,

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

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

Answer 3

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 $\log_2$:

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