Question

e^x-1-5=5
a.1+ln2+ln5
b.1+ln2-ln5
c.1-ln2+ln5
d.ln2+ln5

Answer 1

\(\large \bf e^{x-1}-5=5 ?\)

Answer 2

Add both sides by 5:
\(e^{x-1} = 10\)

Take natural log of both sides:
\(x-1 = \ln(10) = \ln (5)+\ln(2)\)

Add both sides by 1:
\(\boxed{x = \ln (5)+\ln(2)+1}\)

Answer 3

Thank you so much!