# If log(base 3) 4 is approximately 1.2619, evaluate log(base 3) .25. Give your answer correct to four decimals.

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`log_3(0.25)`

`=log_3(25/100)`

`=log_3(25)-log_3(100)` (since `log_a(m/n)` =`log_am-log_an` )

`=log_3(25)-log_3(4*25)`

`=log_3(25)-{log_3(4)+log_3(25)}` (since `log_a(m*n)=log_am+log_an ` )

`=log_3(25)-log_3(4)-log_3(25)`

`=-log_3(4)`

`=-1.2619.` (given)

**Therefore, the answer is -1.2619.**

**Sources:**

`log_3 4=1.2619`

`log_3 .25=log_3 (25/100)`

`=log_3 (1/4)=log_3 4^(-1)=-1xxlog_3 4`

`=-1.2619` =-1.2619

because

`log_3(x/y)=log_3 x-log_3 y`

`log_3 x^m=mlog_3 x`

`1/x^n=x^(-n)`

**Answer is -1.2619**