Solving for the Exponent Value in an Equation

Question

I need to solve the following equation:

$1.024^t = 2$

I cannot use logarithms to solve for $t$ because we haven't formally learned it in class yet. Is there another way to solve for $t$? Any help will be greatly appreciated.

Answer 1

Is there another way to solve for $t$?

Short Answer: No.

Long Answer: You will not be able to reach an exact answer to this without the usage of logarithms, as the answer is irrational. The concepts of logarithms are not hard to understand.

You can simply take $\log_{1.024}$ on both sides, so $t=\log_{1.024}2$. If $\log_{1.024}2$ seems ugly and you want your answer to be in terms of just $\log$(which is $\log_{10}$), you can use the logarithmic base change rule to get:$$t=\dfrac{\log 2}{\log 1.024}$$

Answer 2

First I’m going to assume that the question is $1024^t = 2$ because that is more likely the problem being asked.

My first step would be to calculate the powers of 2. $2^1 = 2$, $2^2 = 2$ $...$. Until I got $2^{10} = 1024$. Knowing this I would reframe the question as.

$$(2^{10})^t = 2^1$$ $$2^{10t} = 2^1$$ $$ 10t = 1$$ $$ t = \frac{1}{10} $$

If the question is “1.024” then the question gets harder.

$(2^{10}10^{-3})^t = 2^1$

Now I just happen to know that $2^{3.32...} = 10$ but if you didn’t know that then there is no way to work this out without using logs.

$$(2^{10}(2^{3.32...})^{-3})^t = 2^1$$ $$(2^{10}2^{-9.96...})^t = 2^1$$ $$2^{(0.04...)t} = 2^1 $$ $$ (0.04...)t = 1$$ $$ t \approx 25$$

However given that finding the solution requires you to already happen to know the logarithm of 10 base 2 I’m going to suggest you metagame the question you have been given and check if you haven’t made a mistake in your calculations leading up to this point.