if 8^x = y, then what is 2^5x-3 equal to in terms of y?

Given the equation \( 8^x = y \), we need to find the value of \( 2^{5x - 3} \) in terms of \( y \).

First, rewrite \( 8^x = y \) using base 2. Since \( 8 = 2^3 \), we can express 8 as:

\[ 8^x = (2^3)^x \]

Using the properties of exponents, this becomes:

\[ (2^3)^x = 2^{3x} \]

So, we have:

\[ 2^{3x} = y \]

Now, we need to express \( 2^{5x - 3} \) in terms of \( y \). From the equation above, we know:

\[ 2^{3x} = y \]

We can solve for \( 2^{x} \):

\[ 2^{3x} = y \]
\[ 3x = \log_2(y) \]
\[ x = \frac{\log_2(y)}{3} \]

Next, substitute \( x \) into \( 2^{5x - 3} \):

\[ 2^{5x - 3} = 2^{5 \left(\frac{\log_2(y)}{3}\right) - 3} \]

Simplify the exponent:

\[ 2^{5 \left(\frac{\log_2(y)}{3}\right) - 3} = 2^{\frac{5 \log_2(y)}{3} - 3} \]

This can be further simplified using the properties of exponents and logarithms:

\[ 2^{\frac{5 \log_2(y)}{3} - 3} = 2^{\frac{5 \log_2(y)}{3}} \cdot 2^{-3} \]

Rewrite it as:

\[ 2^{\frac{5 \log_2(y)}{3}} \cdot 2^{-3} \]

Since \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \), the expression becomes:

\[ \frac{2^{\frac{5 \log_2(y)}{3}}}{8} \]

Recognize \( 2^{\frac{5 \log_2(y)}{3}} \) as \( y \) raised to the power \( \frac{5}{3} \):

\[ 2^{\frac{5 \log_2(y)}{3}} = y^{\frac{5}{3}} \]

Thus, the final expression is:

\[ 2^{5x - 3} = \frac{y^{\frac{5}{3}}}{8} \]

So, \( 2^{5x - 3} \) in terms of \( y \) is:

\[ 2^{5x - 3} = \frac{y^{\frac{5}{3}}}{8} \]

Would you like more details or have any questions?

Here are some related questions for further exploration:
1. How do you derive the change of base formula for logarithms?
2. What are some common properties of exponents?
3. Can you prove the properties of logarithms used in this solution?
4. How do you convert from one logarithm base to another?
5. What are the steps to simplify \( \log_b(a^c) \)?
6. How do you solve exponential equations involving different bases?
7. What is the relationship between exponential and logarithmic functions?
8. How can you apply these principles to solve real-world problems involving exponential growth?

**Tip:** When solving problems involving exponential functions, it is often helpful to express all terms using the same base to simplify calculations and identify relationships more easily.

Explore how to solve the expression 2^5x - 3 in terms of 8^x = y with a detailed mathematical explanation. This problem involves exponential functions, logarithms, and properties of exponents. Suitable for advanced high school students.

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