$$ \mathrm{P}\left(8 ; 1 \frac{1}{2}\right) $$ is a point on the graph of $$ f(x)=\log _{a} x $$ (1) Show that $$ a=4 $$.

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To show that $$ a=4 $$, we need to find the value of $$ a $$ that satisfies the equation $$ f(x) = \log_a x $$ for the given point $$ (8, 1\frac{1}{2}) $$.

Given that $$ f(x) = \log_a x $$, we can substitute the coordinates of the point $$ (8, 1\frac{1}{2}) $$ into the equation to get:

$$ f(8) = \log_a 8 = 1\frac{1}{2} $$

Now, we need to solve for $$ a $$ in the equation $$ \log_a 8 = 1\frac{1}{2} $$.

Let's solve for $$ a $$ step by step.
Solve the equation by following steps:
- step0: Solve for $$a$$:
  $$\log_{a}{\left(8\right)}=1\frac{1}{2}$$
- step1: Find the domain:
  $$\log_{a}{\left(8\right)}=1\frac{1}{2},a \in \left(0,1\right)\cup \left(1,+\infty\right)$$
- step2: Convert the expressions:
  $$\log_{a}{\left(8\right)}=\frac{3}{2}$$
- step3: Convert the logarithm into exponential form:
  $$8=a^{\frac{3}{2}}$$
- step4: Swap the sides:
  $$a^{\frac{3}{2}}=8$$
- step5: Raise both sides to the $$\frac{2}{3}$$-th power$$:$$
  $$\left(a^{\frac{3}{2}}\right)^{\frac{2}{3}}=8^{\frac{2}{3}}$$
- step6: Evaluate the power:
  $$a=4$$
- step7: Check if the solution is in the defined range:
  $$a=4,a \in \left(0,1\right)\cup \left(1,+\infty\right)$$
- step8: Find the intersection:
  $$a=4$$
The value of $$ a $$ that satisfies the equation $$ f(x) = \log_a x $$ for the given point $$ (8, 1\frac{1}{2}) $$ is $$ a = 4 $$.
$$ a = 4 $$

is a point on the graph of
(1) Show that .

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