Question - Solving Logarithmic Equation for x
Solution:
This question is an equation involving logarithms that we need to solve for x:$$ \frac{\log(32)}{\log(x)} + \frac{\log(16)}{\log(x)} = \log(x) $$Firstly, we can combine the terms on the left-hand side using the fact that $$ \log(a) + \log(b) = \log(a \cdot b) $$. However, since these logarithms have the same denominator, we can combine them as follows:$$ \frac{\log(32) + \log(16)}{\log(x)} = \log(x) $$Next, we use the property of logarithms that states $$ \log(a) + \log(b) = \log(ab) $$ to combine the numerators:$$ \frac{\log(32 \cdot 16)}{\log(x)} = \log(x) $$$$ \frac{\log(512)}{\log(x)} = \log(x) $$Now we have a single logarithm in the numerator. We can simplify this further by using the property $$ \log_b(a) = c $$ implies $$ b^c = a $$:Let's equate the left side to the right side. This gives us an equation of the form:$$ \log(512) = \log(x^2) $$Since the logarithms are in the same base and the equation equates one log to another, we can deduce that the arguments must be equal:$$ 512 = x^2 $$Taking the square root of both sides gives us:$$ x = \sqrt{512} $$$$ x = 16 \times \sqrt{2} $$$$ x = 16 \times 1.414... $$ (since $$ \sqrt{2} $$ is approximately 1.414)$$ x \approx 22.63 $$Therefore, the value of $$ x \approx 22.63 $$. Please note this value has been rounded to two decimal places.
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