https://file.aiquickdraw.com/mathbot/file/1731833029348j4y5z22e.jpg

Let's solve the two equations step by step:

### First Equation:
\[ 2 \log y = \log 32 - \log x \]

#### Step 1: Simplify using logarithmic properties
Using the properties of logarithms:
- \( a \log b = \log b^a \)
- \( \log a - \log b = \log \frac{a}{b} \)

Rewriting:
\[ \log y^2 = \log \frac{32}{x} \]

Since the logs are equal, we can equate their arguments:
\[ y^2 = \frac{32}{x} \]

#### Step 2: Solve for \( y \):
\[ y = \sqrt{\frac{32}{x}} = \frac{\sqrt{32}}{\sqrt{x}} = \frac{4\sqrt{2}}{\sqrt{x}} \]

---

### Second Equation:
\[ 4^x = 16^x \]

#### Step 1: Express \( 16^x \) as \( (4^2)^x \):
\[ 4^x = (4^2)^x \]
\[ 4^x = 4^{2x} \]

#### Step 2: Simplify the exponents
Since the bases are the same, we equate the exponents:
\[ x = 2x \]

#### Step 3: Solve for \( x \):
\[ x - 2x = 0 \]
\[ -x = 0 \]
\[ x = 0 \]

However, \( x = 0 \) is not valid because it would make the logarithms undefined in the first equation (division by zero in \( \log x \)).

Thus, no solution exists for the second equation that satisfies \( x > 0 \).

Solve the given equations:
1. 2 log y = log 32 - log x
2. 4^x = 16^x

Step-by-step solution to solve logarithmic and exponential equations: 2 log y = log 32 - log x and 4^x = 16^x. Explore the use of logarithmic properties and laws of exponents in this advanced math problem. Ideal for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation