Solve for y: ln (y-1)- ln 2 = x + ln x. Need somebody to double-check answer.?

Question

I got y = (e ^ 2x + 0.5) *2. Is that correct?

Answer 1

I have written out the steps for you to follow.  You need the basic rules of logarithms, and also the knowledge that e^(ln(z)) = z.

ln(y - 1) - ln(2) = x + ln(x)   ...add ln(2) to both sides

ln(y - 1) = x + ln(x) + ln(2)   ...ln(a) + ln(b) = ln(ab)

ln(y - 1) = x + ln(2x)   ...log_a(c)=b is the same as a^b=c

e^(x + ln(2x)) = y - 1   ...add 1 to both sides

y = e^(x + ln(2x)) + 1   ...a^(b+c)=(a^b)(a^c)

y = e^(x)e^(ln(2x)) + 1

e^(ln(2x)) = z

Since a^b = c is the same as log_a(c) = b this becomes

log_e(z) = ln(2x)

ln(z) = ln(2x)

z = 2x

Then e^(ln(2x) = 2x.

Substitute this fact below.

y = e^(x)e^(ln(2x)) + 1

y = e^(x)2x + 1

y = 2xe^(x) + 1

Hope this helps you!

Answer 2

For this problem you use the property of logarithms, which are :

log(ab) = log(a) + log(b)    Multiplication Property

log(a/b) = log(a) - log(b)    Division Property

log(x^n) = n * log(x)          Power Rule

Also a useful fact to remember is ln (e^x) = x.

I got y = 2x*(e^x) + 1 as an answer.

Make sure you use the Logarithm properties above.