How can I solve for x?

Question

ln(9-3x)= -3

How can I solve for "x"?

also, ln(ln(x))=1??

Answer 1

Question 1
ln (9 - 3x) = - 3
9 - 3x = e^(-3)
9 - 3x = 1 / e ³
3x = 9 - 1 / e ³
x = 3 - 1 / 3 e ³

Question 2
ln (ln x) = 1
ln x = e
x = e^e

Answer 2

exp (ln (9 - 3x) = exp (-3)

9 - 3x = 0.049787

3x = 8.95
x = 2.983404311

exp [ln(ln x)] = exp (1)

ln x = 2.71828128

exp (ln x) = exp (2.71828128)

x = 15.15425393

substitute x into your second formula and you get 1 = 1

Answer 3

ln(9 - 3x) = -3
Exponentiate both sides
9 - 3x = e^(-3)
-3x = e^(-3) - 9
x = 3 - (e^(-3))/3

ln(ln(x)) = 1
Exponentiate both sides
ln(x) = e
Exponentiate again
x = e^e

Answer 4

multiply both sides by power of e

e^(ln(9-3x)) = e^(-3)

9-3x = e^(-3)

-3x = e^(-3) - 9

x = (e^(-3) - 9) / (-3)



ln(ln(x))=1

multiply power of e both sides TWICE, for each ln

e^ln(ln(x)) = e^1 is just e

e^(ln(x)) = e^e

x = e^e

Answer 5

ln (9 - 3x) = -3 // Take exp of both sides

9 - 3x = e^-3

-3x = e^-3 - 9 // divide by -3

x = 3 - (e^-3)/3

-------------------------------------
ln(ln(x)) = 1

ln(x) = e

x = e^e

Answer 6

U can substitute 2.303 log in place of ln....So u'll get ur answer
i.e.,
2.303 log(9-3x) = -3
log9 - log 3x = - 3/ 2.303