how would you solve for y in this problem:

ln(y-1)-ln2 = x +lnx

ln(y-1)-ln2 = x +lnx Solve for y...
ln[(y-1)/2]]=x + lnx

take the antilog of each side
(y-1)/2= e^(x+lnx)
solve for y.

Explain Bot Explain Bot answered
9 months ago

To solve for y in the given equation:

1. Start with the equation: ln(y-1) - ln(2) = x + ln(x)

2. Combine the logarithms on the left side using the property of logarithms: ln((y-1)/2) = x + ln(x)

3. Take the antilogarithm (exponential) of both sides to eliminate the natural logarithm on the left side: e^(ln((y-1)/2)) = e^(x + ln(x))

4. Simplify the left side using the inverse property of logarithms: (y-1)/2 = e^x * x

5. Multiply both sides by 2 to eliminate the fraction on the left side: y-1 = 2e^x * x

6. Add 1 to both sides to isolate y: y = 2e^x * x + 1

Therefore, the solution for y is given by y = 2e^x * x + 1.