# (1) `log_3 (36)-log_3(4/3)` (2) 2(5)^4x-5=30^x+2 (3) log base 2 ( 14)+log base2 (4/7)

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### 2 Answers

`log_3 36-log_3(4/3)`

To simplify, apply the quotient property of logarithm which is `log_b(M/N)= log_bM-log_bN` .

`=log_3 36 - (log_3 4 - log_3 3)`

`=log_3 36 - log_3 4 + log_3 3`

Then, factor 36.

`=log_3 (4 * 9) - log_3 4 + log_3 3`

Also, apply the product property of logarithm `log_b(M*N)= log_bM + log_b N` .

`=log_3 4 + log_3 9 - log_3 4 + log_3 3`

Then, combine like terms.

`= log_3 4 - log_3 4 + log_3 9 + log_3 3`

`= log_3 9 + log_3 3`

Since `9 = 3^2` , then:

`=log_ 3^2 + log_3 3`

Apply the property `log_b a^m = m log_b a` .

`= 2 log_3 3 + log_ 3 3`

Note that if the base and argument of the logarithm are the same it simplifies to 1 `( log_b b = 1)` .

`= 2(1)+ 1`

`=2+1`

`=3`

**Hence, `log_3 36-log_3(4/3) = 3` .**

(For question #2 and #3, please post it separately in Homework help.)

`A)1-log_3(36)-log_3(4/3)= 1-log_3(3^2 xx 2^2)-(log_3(2^2)-log_3(3))=`

`=1-(2log_3 3 +2log_3 2)-(2log_3 2 -1)=`

`=1-2-2log_3 2-2log_3 2 +1=`

`=-4log_3 2`

`B)2-2xx 5^(4x-5)=30^x+2`

`2-2 xx 5^(4x-5)=2^x xx 3^x xx5^x +2=`

`2xx 5^(4x-5)=2^x xx3^x xx5^x=`

`2^(x-1) xx 3^x xx 5^(5-3x)=1`

now using logaritms:

`(x-1)log 2 +x log 3 + (5-3x) log 5 =0`

`x(log 3+log 2-3log 5)+5log 5-log 2 = 0`

`x=(5log 5-log 2)/(log 3 + log 2 -3 log 5)`

`C)` `3-log_2 14 +log_2(4/7)=3-log_2(7xx2)+log_2 4-log_2 7=`

`=3-(log_2 7+ log_2 2)+2log_2 2-log_2 7=`

`=3-log_2 7 -1 +2-log_2 7=2(2-log_2 7)`