Question

lesson 26 - solving logarithmic equations
score: 50/100 answered: 11/20
question 12
solve for x:
\\(\frac{9}{2}\log_{4}x = -9\\)
\\(x = \square\\)
question help: message instructor

Explanation:

Step1: Isolate the logarithmic term

Divide both sides of the equation \(\frac{9}{2}\log_{4}x = -9\) by \(\frac{9}{2}\).
The formula for dividing both sides by a non - zero number \(a\) in an equation \(ax = b\) is \(x=\frac{b}{a}\). Here, \(a = \frac{9}{2}\) and \(b=-9\), so \(\log_{4}x=\frac{-9}{\frac{9}{2}}\).
Simplify \(\frac{-9}{\frac{9}{2}}\): \(\frac{-9}{\frac{9}{2}}=-9\times\frac{2}{9}=-2\). So we get \(\log_{4}x=-2\).

Step2: Convert logarithmic to exponential form

Recall the definition of logarithms: if \(\log_{a}y = z\), then \(y = a^{z}\) (where \(a>0,a
eq1,y > 0\)).
For the equation \(\log_{4}x=-2\), using the above definition with \(a = 4\), \(z=-2\) and \(y = x\), we can rewrite it as \(x = 4^{-2}\).

Step3: Simplify the exponential expression

Recall the rule of exponents: \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a real number).
For \(a = 4\) and \(n = 2\), \(4^{-2}=\frac{1}{4^{2}}\). And \(4^{2}=16\), so \(4^{-2}=\frac{1}{16}\).

Answer:

\(\frac{1}{16}\)