+118724 I'm not quite sure what you mean.
$${\mathtt{1.5}}{\mathtt{\,\times\,}}{\mathtt{1.3}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{0.75}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{\,-\,}}{\frac{{\mathtt{43}}}{{\mathtt{160}}}} = -{\mathtt{0.268\: \!75}}$$
And you cannot find the log of a negative number - they don't exist.
+2354 I can interpret this in multiple ways but all have an invalid answer.
I'm going to interpret it as
$$y = ^4log(1.5*1.3*0.5*0.75-1)$$
which is the log {base4} of (1.5*1.3*0.5*0.75-1)
(and y is your answer)
We can rewrite this to $$4^y = 1.5*1.3*0.5*0.75-1$$
Now since $$1.5*1.3*0.5*0.75-1 \leq 0$$ and $$4^y \geq 0$$ for any y
There exist no y for which this equality holds.
Therefore you equation has no answer.
Instead of base 4 I can interpret it as base 10 or a natural logarithm but it still won't give me an answer.
Perhaps you want to check whether you wrote down the question correctly 
Reinout
