Question 5

If $$5^{2x} \times 5^{x+3} \times 5^{5-x} = (\sqrt{5})^{20}$$, then the value of $$x$$ is

Solution

Adding the exponents on the left gives $$5^{2x + x + 3 + 5 - x} = 5^{2x + 8}$$. The right side is $$(\sqrt{5})^{20} = 5^{10}$$. Equating exponents, $$2x + 8 = 10$$, so $$x = 1$$.

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