Solving Logarithmic Equation for x
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.
