LOGARITHM Problem For Fun

Assume that $\log_\alpha X$ has a real solution. Using logarithm identities: $\begin{aligned} \log_\alpha X - \log_\alpha (X-\beta) &= -1 \\ \log_\alpha \left(\frac{X}{X-\beta}\right) &= -1 \\ \frac{X}{X-\beta} &= \alpha^{-1} \\ \frac{X}{X-\beta} &= \frac{1}{\alpha} \end{aligned}$ Since $\beta>0$ , $X>X-\beta$ and, therefore, $\frac{X}{X-\beta}>1$ . But $\alpha>1$ , so $\frac{1}{\alpha}<1$ , a contradiction.

Thus there are no real solutions, regardless of the values of $\alpha$ and $\beta$ . The answer is None .

If the given equation is true, then $(x-B)>x$ , for A>1.

That would mean B<0, which contradicts the given statement.

We can Simplify the expression above to X= - $\beta$ /( $\alpha -1)$ Since alpha and beta are positive X must be negative .Let $log_\alpha X = n$ .Then there will be no real number that satisfy n because no positive number to the power of a real number is a negative.