Solve the following equations. Leave answers in simplest terms a)e^(3x-9)=8. b) ln(2y+5)=2+ln(4-y)

a) Using log rules, a^x=b becomes log(a)b=x. If we take ln of both sides, we get ln(e^(3x-9))=ln8. ln(e) =1, so we just get 3x-9=ln8. Now we can simply manipulate this to get x=(9+ln8)/3=3+(1/3)ln8. Another log rule is that alogb = log a^b, so we can simplify this further to get x=3+ln8^1/3=3+ln2. b)first, take ln(4-y) to the LHS to get ln(2y+5)-ln(4-y)=2. Using another log rule, lna-lnb=ln(a/b), so we get ln((2y+5)/(4-y))=2. We can take the exponential of both sides to get rid of the ln, and we get (2y+5)/(4-y)=e^2. Now we need to get y on its own. To do this, we multiply 4-y up to get 2y+5=4e^2-e^2y. Now, we can move all the y's on one side, and all the terms without y to the other: 2y+e^2y=4e^2-5. We can factor y out: y(2+e^2)=4e^2-5, and divide by this bracket to get y on its own and the answer: y=(4e^2-5)/(2+e^2).