# The logarithm `log_(4b)x^4 = y` , what is `log_(b)x^12` in terms of y

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### 3 Answers

`log_ab = 1/log_ba`

`log_ab^n = nlog_ab = n/log_ba`

`y = log_(4b)x^4`

`y = 4/log_x4b = 4/(log_x4+log_xb)`

`log_bx^12`

`= 12/log_xb`

Let P = `log_bx`

`y = 4/(log_x4+log_xb)`

`y = 4/(log_x4+P)`

`P = 4/y-log_x4`

`log_bx^12`

`= 12/P`

`= 12/[4/y-log_x4]`

`= 12y/[4-ylog_x4]`

`log_bx^12 = (12y)/[4-ylog_x4]`

**Sources:**

It is given that `log_(4b) x^4 = y` . Using the property of logarithm `log_b a = (log_x a)/(log_x b)`

`y = log_(4b) x^4`

=> `y = (log_b x^4)/(log_b 4b)`

`log_b x^12 = log_b (x^4*x^4*(x^4)^(1/2))`

=> `log_b x^4 + log_b x^4 + (1/2)*log_b x^4`

=> `log_b(4b)*(y + y + y/2)`

=> `(log_b 4 + 1)((5y)/2)`

It is not possible to express `log_b x^12` only in terms of y, a constant term has to be included.

**The logarithm `log_b x^12` can be expressed as `2.5*y*(1 + log_b 4)` **

log(4b)x^4 = y by data.

So x^4 = (4b)^y by definition of logarithms.

Therefore x^12 = (x^4)^3 =((4b)^y)^3 = 4^(3y)*b^(3y)

Therefore, log(b)x^12 = log(b){4^(3y)*b^(3y)}

log(b)x^12 = log(b){4^(3y)}+log(b){b^(3y)}

log(b)x^12 = 3ylog(b)4 + 3yl, as logxy = logx+logy, logx^y = ylogx and log(b)b^x = x.

**Therefore log(b)x^12 = 3y{log(b)4 + 1}**