Understanding the Exponential Equation Problem

We are given a problem involving exponential equations and a relationship between the bases. The problem states that we have three quantities related by powers: $\(a^x = b^y = c^z\)$. Additionally, there is a geometric mean relationship between the bases $a$, $b$, and $c$, given by $\(b^2 = ac\)$. Our goal is to find the value of $\(y\)$ in terms of $\(x\)$ and $\(z\)$ by solving this exponential equation.

Setting up the Equations

Let's introduce a constant, say $\(k\)$, to represent the common value of the exponential equations:

$\(a^x = b^y = c^z = k\)

From this, we can express $\(a\)$, $\(b\)$, and $\(c\)$ in terms of $\(k\)$ and their respective exponents. Using the definition of logarithms or fractional exponents:

• $\(a^x = k \implies a = k^{1/x}\)$
• $\(b^y = k \implies b = k^{1/y}\)$
• $\(c^z = k \implies c = k^{1/z}\)$

These expressions will be crucial in solving this algebra problem.

Applying the Geometric Mean Relationship

The second piece of information given is the relationship between the bases: $\(b^2 = ac\)$. Now, we substitute the expressions for $\(a\)$, $\(b\)$, and $\(c\)$ that we found in terms of $\(k\)$ into this equation. This is a key step in solving for $\(y\)$.

Substitute $\(a = k^{1/x}\)$, $\(b = k^{1/y}\)$, and $\(c = k^{1/z}\)$ into $\(b^2 = ac\)$:

$\(\left(k^{1/y}\right)^2 = \left(k^{1/x}\right) \left(k^{1/z}\right)\)

Using Power Rules for Solving the Exponential Equation

Now we simplify the equation using standard power rules. Recall that $\((p^m)^n = p^{mn}\)$ and $\(p^m \cdot p^n = p^{m+n}\)$.

Applying the power rules to our equation:

• The left side becomes: $\(\left(k^{1/y}\right)^2 = k^{2 \cdot (1/y)} = k^{2/y}\)$
• The right side becomes: $\(\left(k^{1/x}\right) \left(k^{1/z}\right) = k^{1/x + 1/z}\)$

So the equation becomes:

$\(k^{2/y} = k^{1/x + 1/z}\)

Since the bases are equal ($\(k\)$) and they are positive (assuming $\(a, b, c\)$ are positive for real exponents), the exponents must be equal. This allows us to transition from an exponential equation to a simple algebraic one.

$\(\frac{2}{y} = \frac{1}{x} + \frac{1}{z}\)

This is a common form encountered when solving for y in such mathematical equations.

Solving for y

We now have a simple algebraic equation relating $\(x\)$, $\(y\)$, and $\(z\)$. We need to isolate $\(y\)$. First, combine the terms on the right side by finding a common denominator:

$\(\frac{1}{x} + \frac{1}{z} = \frac{z}{xz} + \frac{x}{xz} = \frac{z+x}{xz}\)

So the equation is:

$\(\frac{2}{y} = \frac{z+x}{xz}\)

To solve for $\(y\)$, we can take the reciprocal of both sides:

$\(\frac{y}{2} = \frac{xz}{z+x}\)

Finally, multiply both sides by 2 to get the expression for $\(y\)$:

$\(y = \frac{2xz}{z+x}\)

This gives us the value of $\(y\)$ in terms of $\(x\)$ and $\(z\)$ by successfully navigating the given mathematical equations and relationships. This method provides a clear approach to solving for y in this particular algebra problem by using the properties of exponents and algebraic manipulation.