Solving an Exponential Equation Algebraically and Graphically

[Illvoices]

Solve the equation e^3-2x=4 algebraically and graphically.
e^3-2x=4
ln e^3-2x=ln4
3-2=ln4
-2x=-3 +4
x= 1/2(3-4ln) aproximately 0.807

so i would want to know how did the 1/2 get there in the last equation? i couldn't figure this one out would anyone answer this and tell me why'd the book didn't show it.

 

[JeffM]

Solve the equation e^3-2x=4 algebraically and graphically.
e^3-2x=4
ln e^3-2x=ln4 Good
3-2=ln4 What happened to x?
-2x=-3 +4 How did ln(4) become 4?
x= 1/2(3-4ln) aproximately 0.807 4 * ln what? The letters ln stand for a FUNCTION. It must be ln(some expression).

so i would want to know how did the 1/2 get there in the last equation? i couldn't figure this one out would anyone answer this and tell me why'd the book didn't show it. Probably the book did not show it because
if you have - 2x = q, it is fairly simple to multiply both sides of the equation by - (1/2) to get x = - (1/2)q.

Let's do it step by step.

\(\displaystyle e^{(3 - 2x)} = 4 \implies ln \left ( e^{(3 - 2x)} \right ).\) Good so far.

\(\displaystyle ln \left ( e^{(3 - 2x)} \right ) = ln(4) \implies (3 - 2x) * ln(e) = ln(4).\)

\(\displaystyle (3 - 2x) * ln(e) = ln(4) \implies (3 - 2x) * 1 = ln(4).\)

\(\displaystyle (3 - 2x) * 1 = ln(4) \implies 3 - 2x = ln(4).\)

\(\displaystyle 3 - 2x = ln(4) \implies -\ 2x = ln(4) - 3.\)

\(\displaystyle -\ 2x = ln(4) - 3 \implies -\ \left ( \dfrac{1}{2} \right) * (-\ 2x) = -\ \left ( \dfrac{1}{2} \right) * \{ln(4) - 3\}.\)

\(\displaystyle -\ \left ( \dfrac{1}{2} \right) * (-\ 2x) = -\ \left ( \dfrac{1}{2} \right) * \{ln(4) - 3\} \implies x = 1.5 + \left \{-\ \dfrac{1}{2} * ln(4) \right \}.\)

\(\displaystyle x = 1.5 + \left \{-\ \dfrac{1}{2} * ln(4) \right \} = 1.5 + ln \left ( 4^{-(1/2)} \right ) \implies\)

\(\displaystyle x = 1.5 + ln \left ( \left \{ \dfrac{1}{4} \right \}^{(1/2)} \right ) = 1.5 + ln(0.5) \approx 0.807.\)

 

[Illvoices]

ok thank you for the help teach, because of this i was able to understand the problem and solve in my own terms.