# Thread: exponential equation (With pictures!)

1. ## exponential equation (With pictures!)

This should just about sum it up.

Thank you interent people!

2. Originally Posted by Vamz
This should just about sum it up.

Thank you interent people!
$\displaystyle f(x) = -2^{x - 1} + \frac{7}{2}$.

To find the $\displaystyle x$ intercept, let $\displaystyle f(x) = 0$.

So $\displaystyle -2^{x - 1} + \frac{7}{2} = 0$

$\displaystyle 2^{x - 1} = \frac{7}{2}$

$\displaystyle 2\cdot 2^{x - 1} = 7$

$\displaystyle 2^{x} = 7$

$\displaystyle \ln{(2^x)} = \ln{7}$

$\displaystyle x\ln{2} = \ln{7}$

$\displaystyle x = \frac{\ln{7}}{\ln{2}}$.

3. Originally Posted by Prove It
$\displaystyle f(x) = -2^{x - 1} + \frac{7}{2}$.

To find the $\displaystyle x$ intercept, let $\displaystyle f(x) = 0$.

So $\displaystyle -2^{x - 1} + \frac{7}{2} = 0$

$\displaystyle 2^{x - 1} = \frac{7}{2}$

$\displaystyle 2\cdot 2^{x - 1} = 7$

$\displaystyle 2^{x} = 7$ what happeend to ^(x-1) .. and that other 2?

$\displaystyle \ln{(2^x)} = \ln{7}$

$\displaystyle x\ln{2} = \ln{7}$

$\displaystyle x = \frac{\ln{7}}{\ln{2}}$.

Your final answer does not equal to ln(7)

what I posted was what my teacher had in the answer key. I am assuming he's right... unless he was wrong!?

4. You should know that

$\displaystyle a^m \cdot a^n = a^{m + n}$.

Here you have

$\displaystyle 2^1 \cdot 2^{x - 1} = 2^{1 + x - 1} = 2^x$.

And your teacher is wrong.

Either s/he meant $\displaystyle \frac{\ln{7}}{\ln{2}}$ or s/he meant $\displaystyle \log_2{7}$.

5. Thank you.

You sir, are a smart cookie. And I'd give you one, but I wouldn't know where to begin contacting you.

case closed.

6. Originally Posted by Vamz
Thank you.

You sir, are a smart cookie. And I'd give you one, but ...
Hallo,

a personal remark: Before you start throwing cookies around you could instead press the $\displaystyle \boxed{\text{Thanks}}$-button. That's the kind of food we are living on here .

EB