Use logarithms to solve log 8 / log x < 1.
I understand that there are 2 cases: when x is negative or positive.
I don't understand how it affects the answer though.
log8<logx if logx is positive
10^8<x
Log8>logx if logx is negative.
10^8>x if x is less one
When dealing with logarithmic inequalities, it is important to consider the restrictions on the domain of the logarithmic function. In this case, we have the inequality:
log 8 / log x < 1
To solve this inequality, we need to consider two cases: when x is positive and when x is negative.
Case 1: x is positive
In this case, we can assume that both the numerator log 8 and the denominator log x are positive because logarithms of positive numbers are always positive. Therefore, we can rewrite the inequality as:
log 8 < log x
Since logarithmic functions are increasing, this means that the base 10 logarithm of 8 is less than the base 10 logarithm of x. By applying the inverse function, we have:
8 < x
So, for x values greater than 8, the inequality holds in this positive case.
Case 2: x is negative
When x is negative, the logarithm of x is undefined because we can only take the logarithm of positive numbers. Therefore, in this case, the inequality is not valid.
To summarize, the solution to the inequality log 8 / log x < 1 is x > 8 for positive x values. The inequality is not valid for negative x values because the logarithm of a negative number is undefined.