# Answer to Question #16956 in Calculus for hsd

Question #16956

Solve each equation for x. (a) 5^(x−5)=2 (b) lnx+ln(x−1)=1?

Expert's answer

(a) 5^(x−5)=2 then log_5 (5^(x−5) )=log_5 (2)

x-5=log_5 (2) x=5+log_5

(2)

(b) ln(x)+ln(x-1)=1 we have that x>1 from definition of

log

ln(x)+ln(x-1)=ln( x*(x-1) )=1

e^ln( x*(x-1) )

=e^1

x^2-x-e=0

D=1+4e

x1=(1+sqrt(1+4e) )/2

x2=(1-sqrt(1+4e)

)/2

x2<1 so it cant be a root so x1=(1+sqrt(1+4e) )/2 is a solution

x-5=log_5 (2) x=5+log_5

(2)

(b) ln(x)+ln(x-1)=1 we have that x>1 from definition of

log

ln(x)+ln(x-1)=ln( x*(x-1) )=1

e^ln( x*(x-1) )

=e^1

x^2-x-e=0

D=1+4e

x1=(1+sqrt(1+4e) )/2

x2=(1-sqrt(1+4e)

)/2

x2<1 so it cant be a root so x1=(1+sqrt(1+4e) )/2 is a solution

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