# Find the exponential value of a number.

I'm wondering if there is a way to add two numbers with the same base but different exponents without converting the base of the numbers. For example lets say your adding 2^2 and 2^4 the value of 2^2 is 4 and the value of 2^4 is 16 so the total of the two numbers is 20. The closest I could get my calculator to finding a base of 2 with a value of 20 was 2^4.3219281 which equaled 20.000000071. So I have a few questions.

1. Can 2 be raised to a power that's exactly equal to 20?

2. I know that with division and multiplication you can just add or subtract the exponents. Are there any similar tricks for adding or subtracting values with exponents?

3. Is there anyway to arrive at the result, or a similar result of 2^4.3219281 without changing the base in the process?

• Are you familiar with logarithms? – Matthew Conroy Nov 8 '16 at 20:27
• I'm not but I would be willing to read up on them if it would help. – Kahless Nov 8 '16 at 20:42
• Well. $b^x+b^y=b^x (1+b^{y-x})$. But that is usually not much help.. – fleablood Nov 8 '16 at 21:02

1. There is a particular function just designed for solving problems of the form $b^x=a$. The solution to said problem is given by$$x=\log_b(a)=\frac{\log(a)}{\log(b)}$$where we use the second form if your calculator doesn't specify the base of the logarithm. For example,$$2^x=20\implies x=\log_2(20)$$

2. No, definitely no simple tricks. $\ddot\frown$

3. As to calculating what $\log_2(20)$ is without a calculator, one can implement the following algorithm:$$2^x=20=5\times2^2=2^{\log_2(5)}2^2=2^{2+\log_2(5)}$$$$x=2+\log_2(5)$$So the problem now is to find $\log_2(5)$, which is much easier to approximate. Clearly, it's larger than $2$, so maybe $\log_2(5)\approx2.3$, giving us $x=4.3$

You seem to be looking for an exponent $f(a,b)$ with the property that $$2^{f(a,b)} =2^a+2^b$$ The simple answer is that you want to take $$f(a,b) = \log_2(2^a+2^b)$$ But that's just the definition of the logarithm ($x^y=z\iff y=\log_x z$). Other than this, there's not any real way to do the conversion.

In this specific case, you used the number $\log_2 20\approx 4.32193$.

There is no "trick" for $\log_2 (2^a + 2^b)$, other than

$$\log_c (c^a + c^b) = \log_c (c^a (1 + c^{b-a})) = a + \log_c(1 + c^{b-a})$$

In particular $\log_2(20) = 2 + \log_2(5)$.

$\log_2(5)$ is a transcendental number by the Gelfond-Schneider theorem.