Logarithms - Logarithms Properties-advanced

Evaluate the expression: $\frac{1}{\log_2 30} + \frac{1}{\log_3 30} + \frac{1}{\log_5 30}$

Using the reciprocal property $\frac{1}{\log_a b} = \log_b a$, we can rewrite the terms: $\log_{30} 2 + \log_{30} 3 + \log_{30} 5$ Using the product rule $\log_a m + \log_a n + \log_a p = \log_a (m \times n \times p)$: $\log_{30} (2 \times 3 \times 5)$ $\log_{30} 30$ Since $\log_a a = 1$, the result is: $1$

The reciprocal property allows us to bring the terms to a common base of $30$. Once the bases are the same, we apply the product rule of logarithms.

Solve for $x$: $\log_2 x + \log_4 x + \log_{16} x = \frac{7}{4}$

Express all logarithms with base $2$ using the rule $\log_{a^k} m = \frac{1}{k} \log_a m$: $\log_2 x + \log_{2^2} x + \log_{2^4} x = \frac{7}{4}$ $\log_2 x + \frac{1}{2} \log_2 x + \frac{1}{4} \log_2 x = \frac{7}{4}$ Factor out $\log_2 x$: $\log_2 x \left(1 + \frac{1}{2} + \frac{1}{4}\right) = \frac{7}{4}$ $\log_2 x \left(\frac{4+2+1}{4}\right) = \frac{7}{4}$ $\log_2 x \left(\frac{7}{4}\right) = \frac{7}{4}$ Dividing both sides by $\frac{7}{4}$: $\log_2 x = 1$ Converting to exponential form: $x = 2^1 = 2$

To solve logarithmic equations with different bases that are powers of each other, convert all terms to the smallest base (base $2$ in this case) using the base power property.

If $x^2 + y^2 = 7xy$, prove that $2 \log\left(\frac{x+y}{3}\right) = \log x + \log y$

Given $x^2 + y^2 = 7xy$. Add $2xy$ to both sides to complete the square: $x^2 + y^2 + 2xy = 7xy + 2xy$ $(x+y)^2 = 9xy$ Divide both sides by $9$: $\frac{(x+y)^2}{9} = xy$ $\left(\frac{x+y}{3}\right)^2 = xy$ Taking logarithm on both sides: $\log \left(\frac{x+y}{3}\right)^2 = \log (xy)$ Apply the power rule on the left and the product rule on the right: $2 \log \left(\frac{x+y}{3}\right) = \log x + \log y$

We use algebraic manipulation to create a perfect square, then apply logarithmic properties (power and product rules) to reach the required identity.