Question

solve the following logarithm problem for the positive solution for ( x ).( log_{343} x = -\frac{2}{3} )

Explanation:

Step1: Recall the logarithmic definition

The logarithmic equation \(\log_{b} a = c\) is equivalent to \(b^{c}=a\) (by the definition of logarithms, where \(b>0,b
eq1\) and \(a > 0\)). Here, \(b = 343\), \(c=-\frac{2}{3}\) and we need to find \(a=x\). So we can rewrite the equation as \(x = 343^{-\frac{2}{3}}\).

Step2: Simplify the exponent

We know that \(343=7^{3}\) (since \(7\times7\times7 = 343\)). Substitute \(343 = 7^{3}\) into the equation: \(x=(7^{3})^{-\frac{2}{3}}\).
Using the power - of - a - power rule \((a^{m})^{n}=a^{m\times n}\), we have \(x = 7^{3\times(-\frac{2}{3})}\).
Simplify the exponent: \(3\times(-\frac{2}{3})=-2\). So \(x = 7^{-2}\).

Step3: Evaluate the negative exponent

Using the rule \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a real number), for \(a = 7\) and \(n = 2\), we get \(x=\frac{1}{7^{2}}\).
Since \(7^{2}=49\), then \(x=\frac{1}{49}\).

Answer:

\(\frac{1}{49}\)