# 1. Logs: Evaluate (round to two decimals where neccassary): a) Log(base 8) 4 + Log(base 8) 16b) 10 ^(logx) base=very smallc) Express as a single Log in simplified form: Log (base 2) (x+8) +...

1. Logs: Evaluate (round to two decimals where neccassary):

a) Log(base 8) 4 + Log(base 8) 16

b) 10 ^(logx)

base=very small

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)

*print*Print*list*Cite

### 1 Answer

(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.**

**Sources:**