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Question 4
Consider the equation $$\log_5(x - 2) = 2 \log_{25}(2x - 4)$$, where x is a real number.
For how many different values of x does the given equation hold?
Solution
Let's reduce the RHS first :-$$2 \log_{25}(2x - 4)$$ = $$\frac{2}{2}\log_5(2x-4)$$ = $$ \log_{5}(2x - 4)$$
Here we used the formulae $$\log_{a^n}b^m=\frac{m}{n}\log_ab$$
Now LHS=RHS
$$\log_5(x - 2) = \log_{5}(2x - 4)$$
The bases are equal.
So, x-2=2x-4
x=2, but 2 is not in the domain as it will lead to log 0. Hence, there are no solutions for this.
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