# Calculate logarithm base 30 of 8 if lg5=a and lg3=b

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According to the logarithm change of base rule,

`log_(b)(x)=(log_(c)(x))/(log_(c)(b))`

lg stands for log with base 10. So,

`log_(30)8=(log_(10)8)/(log_(10)30)`

`=(log(2)^(3))/(log(2*3*5))`

`=(3log2)/(log2+log3+log5)` ...........(i)

Again, log2=log(10/5)=log10-log5 =1-a [ since log5=a....(given)]

Now, putting the values of log2, log3 and log5 in (i) we get:

`log_(30)8` =`(3*(1-a))/(1-a+b+a)`

`=(3-3a)/(1+b)`

**Hence, the required answer is** `(3-3a)/(1+b)`