Question

solve ln(y-1)-ln 2 = x + ln x

Answer 1

solve for y

Answer 2

\( \ln(y-1)- \ln 2 = x + \ln x\)

some starting hints

\( \ln \dfrac{y-1}{ 2} = x + \ln x\)

\(\ln \dfrac{y-1}{ 2} = \ln e^x + \ln x\)

2 more steps!

Answer 3

wait why was x on the right side of the equation put int exponential form? @IrishBoy123

Answer 4

because you can then say that


\( \ln e^x + \ln x\)

\(= \ln x e^x \)

Answer 5

make sense?

Answer 6

so do you then multiply by 2 and then add 1 on each side?

Answer 7

well, you've now got this

\(\ln \dfrac{y-1}{ 2} = \ln x e^x\)

so you can "un-log" it, if that makes sense

Answer 8

how do you do that? do you divide both sides by ln?

Answer 9

nah! just "un-function" them


if \( \ln P = \ln Q\) then \(P = Q\)

Answer 10

ohhhhhh okay so they just automatically cancel out one another

Answer 11

yes!

Answer 12

So

\( \dfrac{y-1}{ 2} = x e^x\)

Answer 13

still needs to be finished off

Answer 14

y = 2xe^x+1

Answer 15

\(\Huge \checkmark\)

Answer 16

YES thanks so much!!

Answer 17

mp!