- #1
ZedCar
- 354
- 1
Homework Statement
Solve for x
0.5y = e^-x
The Attempt at a Solution
I believe the answer is x = ln(2/y)
Would anyone be able to explain the intermediate steps please?
If I was trying it I would have got -ln0.5y = x
ZedCar said:Homework Statement
Solve for x
0.5y = e^-x
The Attempt at a Solution
I believe the answer is x = ln(2/y)
Would anyone be able to explain the intermediate steps please?
If I was trying it I would have got -ln0.5y = x
cbetanco said:If you take the log of either side you get
[itex]ln(y/2)=-x[/itex] or rearranging terms and using [itex]ln(a/b)=ln(a)-ln(b)[/itex] gives [itex]x=-ln(y/2)=-(ln(y)-ln(2))=ln(2/y)[/itex]
Mentallic said:Or more simply,
[tex]a\cdot \ln(b)=\ln(b^a)[/tex]
so
[tex]-\ln(y)=\ln(y^{-1})=\ln(1/y)[/tex]
The first step is to divide both sides of the equation by 0.5. This will give us y = 2*e^-x. Next, we can take the natural log of both sides to eliminate the exponent. This will leave us with ln(y) = ln(2*e^-x). Using the product rule of logarithms, we can simplify this to ln(y) = ln(2) + ln(e^-x). The ln(e^-x) can then be further simplified to -x, leaving us with the equation ln(y) = ln(2) - x. Finally, we can subtract ln(2) from both sides to isolate x. The final answer is x = ln(2) - ln(y).
We divide both sides by 0.5 to isolate the variable y on one side of the equation. This will make it easier to solve for x in the subsequent steps.
The natural log function, denoted as ln, is the inverse of the exponential function e^x. This means that when we take the ln of both sides of the equation, the ln and e^x will cancel each other out, leaving us with just the variable x.
The product rule of logarithms allows us to simplify the expression ln(2*e^-x) to ln(2) + ln(e^-x). This makes it easier to manipulate the logarithmic terms and isolate the variable x.
Yes, it is possible to solve this equation without using logarithms. However, using logarithms is often the most efficient and straightforward method. Other methods may involve using exponent rules or taking the natural log of both sides multiple times, which can be more complex and time-consuming.