Key Concepts and Formulas

• Exponential Properties: $(a^m)^n = a^{mn}$
• Substitution: Replacing a complex term with a simpler variable to simplify the equation.
• Logarithm Definition and Properties: $b^x = N \iff x = \log_b(N)$, $\log_b(m) + \log_b(n) = \log_b(mn)$, $\log_b(b) = 1$

Step 1: Transforming the Equation into a Quadratic Form

The given equation is: $(8)^{2 x}-16 \cdot(8)^x+48=0$ We can rewrite $(8)^{2x}$ as $(8^x)^2$. This is because of the exponential property $(a^m)^n = a^{mn}$. $(8^x)^2 - 16 \cdot (8^x) + 48 = 0$ Why this step? This transformation reveals that the equation is structured like a quadratic equation. By identifying the common term $(8^x)$, we can make a substitution to simplify it into a standard quadratic form, which is easier to solve algebraically.

Step 2: Solving the Quadratic Equation for y

Let $y = (8)^x$. Substituting $y$ into the transformed equation, we get: $y^2 - 16y + 48 = 0$ Why this step? This substitution simplifies the problem significantly, changing it from an exponential equation to a familiar quadratic equation.

Now, we solve this quadratic equation for $y$. We can do this by factoring or by using the quadratic formula. Let's use factoring: We need to find two numbers that multiply to 48 and add up to -16. These numbers are -12 and -4. $(y - 12)(y - 4) = 0$ Why this step? Factoring allows us to find the values of $y$ that satisfy the equation by setting each factor equal to zero.

Setting each factor to zero, we find the possible values for $y$: $y - 12 = 0 \implies y = 12$ $y - 4 = 0 \implies y = 4$ So, the solutions for $y$ are $12$ and $4$.

Step 3: Finding the Solutions for x

Now, we substitute back $y = (8)^x$ to find the values of $x$.

Case 1: $y = 12$ $(8)^x = 12$ Why this step? We use the definition of logarithms to solve for $x$ when it's in the exponent. If $b^x = N$, then $x = \log_b(N)$.

Applying the definition of logarithm with base 8: $x_1 = \log_8(12)$

Case 2: $y = 4$ $(8)^x = 4$ Applying the definition of logarithm with base 8: $x_2 = \log_8(4)$

The two solutions for $x$ are $\log_8(12)$ and $\log_8(4)$.

Step 4: Summing the Solutions

The problem asks for the sum of all solutions. Let's add the two solutions we found: $\text{Sum of solutions} = x_1 + x_2 = \log_8(12) + \log_8(4)$ Why this step? We are directly addressing the question asked in the problem statement.

Step 5: Simplifying the Result

To simplify the sum of logarithms, we use the logarithm property: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$. $\log_8(12) + \log_8(4) = \log_8(12 \times 4)$ $= \log_8(48)$ Why this step? Combining the logarithms into a single term makes it easier to evaluate or match with the given options.

Now, we need to simplify $\log_8(48)$. We can use the property $\log_b(mn) = \log_b(m) + \log_b(n)$. We look for factors of 48 that include the base 8. We know that $48 = 8 \times 6$. $\log_8(48) = \log_8(8 \times 6)$ $= \log_8(8) + \log_8(6)$ Why this step? This separates the expression into a term we can easily evaluate ($\log_8(8)$) and a remaining term.

Using the property $\log_b(b) = 1$: $\log_8(8) = 1$ Substituting this back, we get: $1 + \log_8(6)$

This matches Option (A).

Common Mistakes & Tips

• Remember to substitute back to find the values of the original variable ($x$) after solving for the substituted variable ($y$).
• Always simplify logarithmic expressions using logarithm properties to match the given options.
• Recognize exponential equations that can be transformed into quadratic forms using appropriate substitutions.

Summary

The problem was solved by recognizing the exponential equation as a quadratic in disguise, using substitution ($y = 8^x$) to convert it into a standard quadratic equation ($y^2 - 16y + 48 = 0$), solving the quadratic equation to find $y=12$ and $y=4$, using the definition of logarithms to find the individual solutions for $x$: $x_1 = \log_8(12)$ and $x_2 = \log_8(4)$, and applying logarithm properties to sum these solutions and simplify the result to $1 + \log_8(6)$.

Final Answer

The final answer is $\boxed{1+\log _8(6)}$, which corresponds to option (A).