when solving the exponential equation, 2^2x-1=16 what do I do first? do I convert to a log equation?
I would take the log base 2 of each side.
I think you mean 2^(2x-1) = 16
taking logs would be the hard way of doing it.
Notice that 16 is a power of 2, namely 16 = 2^4
so 2^(2x-1) = 2^4
clearly 2x-1 = 16
2x=17
x = 17/2
<clearly 2x-1 = 16>
I meant to type
clearly 2x-1 = 4
2x=5
x=5/2
To solve the exponential equation 2^(2x-1) = 16, the first step is to isolate the exponential term. Converting the equation to a logarithmic equation is not necessary at this point, as there is already a single exponential term on one side.
Here's how you can proceed:
1. Start by rewriting 16 as a power of 2. Since 2^4 = 16, you can replace 16 in the equation with 2^4, resulting in 2^(2x-1) = 2^4.
2. Now that the bases are equal (both are 2), you can set the exponents equal to each other. So, 2x - 1 = 4.
3. Next, isolate the variable 2x by adding 1 to both sides of the equation. This gives you 2x = 5.
4. Finally, solve for x by dividing both sides of the equation by 2. This yields x = 5/2 or x = 2.5.
Therefore, the solution to the exponential equation 2^(2x-1) = 16 is x = 2.5.