Ln(y - 7) - ln(7) = x + ln(x), find y. ?

Question

not sure how to approach this one. thanks.

Answer 1

ln (y - 7) - ln (7) = x + ln (x)

Use the quotient property of logarithms.

ln ( (y - 7) / 7 ) = x + ln (x)

ln ( (y - 7) / 7 ) - ln (x) = x

ln ( ( (y - 7) / 7 ) / x ) = x

ln ( ( (y - 7) / 7 ) * (1/x) ) = x

ln ( (y - 7) / 7x ) = x

Take base e raised to the power of both sides.

e^( ln ( (y - 7) ) / 7x ) = e^(x)

(y - 7) / 7x = e^(x)

y - 7 = 7x *e^(x)

y = 7xe^(x) + 7

Answer 2

ln(y - 7) - ln(7) = x + ln(x)

[First, move all ln terms to one side]

ln(y - 7) - ln(7) -  ln(x) = x

[Now, use ln properties to combine ln terms]

[Remember division inside the ln can be turned into subtraction outside the ln, and vice versa.]

ln[(y - 7)/(7x)] = x

[next take e^ of both sides]

e^[ln[(y - 7)/(7x)]] = e^x

[e^ln is equal to one]

(y - 7)/(7x) = e^x

[Next, multiply by 7x on both sides]

(y - 7) = (7x)e^x

[Lastly, add 7 to both sides]

y = 7xe^(x) + 7

Answer 3

I found Y its just next to Z

Answer 4

ln(y - 7) - ln(7) = x + ln(x)

ln((y - 7) / 7)) = x + ln(x)   [property of logarithms]

e^(ln((y - 7) / 7))) = e^(x + ln(x))   [introduce e as a base]

(y - 7) / 7 = (e^x)(e^ln(x))   [properties of logarithms and exponents]

(y - 7) / 7 = (e^x) x   [property of natural logarithms]

y - 7 = 7(e^x) x   [algebra]

y = 7(e^x) x + 7   [algebra]

y = 7(xe^x + 1)   [algebra]

Answer 5

-> Use a law of logartithms

ln((y-7)/7)=x+lnx

e^(ln((y-7)/7)=e^(x+lnx)

(y-7)/7=e^(x+lnx)

y-7=7e^(x+lnx)

y=7+7e^(x+lnx)