What are the restrictions on the variable x in the following equation? Why do these restrictions exist?
log(x-2)-log(3x-5)=5
Thanks for your help :-)
log(n) is defined only for n > 0.
So, we must have x-2 > 0 or x>2
and 3x-5 > 0 or x > 5/3
Combining those conditions, we see we must have x>2. That will make both expressions positive.
Since we cannot take Log of zero, x-2
cannot equal 0. In other words, x cannot equal 2. So x can be any real
number except 2:
X < > 2.
Also, 3x - 5 cannot equal 0:
3x-5 = 0,
3x = 5,
X = 5/3.
So x should be less than or greater than 5/3:
X < > 5/3.
To determine the restrictions on the variable x in the equation log(x-2) - log(3x-5) = 5, we need to consider the domain of the logarithmic functions involved.
In general, the domain of a logarithmic function is all positive real numbers. This is because logarithms are only defined for positive values. However, in this equation, we have two logarithmic terms being subtracted from each other.
For the expression log(x-2), we need x-2 to be greater than zero: (x-2) > 0. Solving this inequality gives x > 2.
Similarly, for the expression log(3x-5), we need 3x-5 to be greater than zero: (3x-5) > 0. Solving this inequality gives x > 5/3.
Therefore, the restrictions on the variable x in the equation are x > 2 and x > 5/3. Combining these two restrictions, we can conclude that the variable x must be greater than the greatest common value of the two restrictions, which is x > 2.
These restrictions exist because logarithmic functions are not defined for non-positive values. Since the equation involves subtracting two logarithmic terms, we need to ensure that the values inside both logarithms (x-2 and 3x-5) are positive. By imposing the restrictions x > 2, we ensure that both expressions are positive, thus satisfying the domain requirements for logarithms.