e^x-2 + 4 = 21

I got .8332, is this correct?

How did you get that ?

is your equation e^(x-2) + 4 = 21 ?

then
e^(x-2) = 17
ln both sides
ln[e^(x-2)] = ln 17
(x-2) ln e = ln 17
x - 2 = ln 17
x = ln17 + 2 = 4.8332

Oh, i subtracted 2 instead of adding it, thank you.

To check if your answer is correct, we need to solve the equation e^x - 2 + 4 = 21.

First, let's combine the constants on the left side:

e^x - 2 + 4 = 21
e^x + 2 = 21

Next, let's isolate the exponential term:

e^x = 21 - 2
e^x = 19

To solve for x, we need to take the natural logarithm (ln) of both sides:

ln(e^x) = ln(19)
x = ln(19)

Using a scientific calculator, we can calculate the natural logarithm of 19, which is approximately 2.9444.

So, x ≈ 2.9444.

Now, let's compare your answer of 0.8332 to the actual solution. As you can see, your answer does not match the actual solution. Perhaps there was a calculation error or misunderstanding of the equation. Please check your calculations again.

Therefore, your answer of 0.8332 is not correct.