1. Logs: Evaluate (round to two decimals where neccassary):
a) Log(base 8) 4 + Log(base 8) 16
b) 10 ^(logx)
c) Express as a single Log in simplified form:
Log (base 2) (x+8) + 2log(base 2) (x-3)
In (x^2-y^2) - In (x+y)
(a) Evaluate `log_8 4+log_8 16` Use the property of logs: logA+logB=logAB
Thus `log_8 4+log_8 16=log_8 64=2` since `8^2=64`
So the answer is 2.
(b) Evaluate `10^(logx)` The base of the logarithm is understood to be 10 (the common logarithm). Logarithms and exponentials to the same base are inverses of each other (undo functions) so `10^(logx)=x`
The answer is x.
(c) Simplify `log_2 (x+8)+2log_2 (x-3)`
First use the property of logs: `alogb=logb^a`
Thus `log_2 (x+8)+2log_2 (x-3)=log_2 (x+8)+log_2 (x-3)^2`
Now use logA+logB=logAB:
`=log_2 (x+8)(x-3)^2` which is the answer.
(d) Simplify `ln (x^2-y^2)-ln (x+y)`
Use the property lnA-lnB=ln (A/B)
Then `ln(x^2-y^2) - ln(x+y)=ln ((x^2-y^2)/(x+y))`
Factoring the numerator and cancelling like terms leaves:
`=ln(x-y)` which is the answer.