https://file.aiquickdraw.com/mathbot/file/1726403677705usmq6esz.jpg

The given simultaneous equations from the image are:

\[
\frac{5^x}{25^{3y-2}} = 1
\]
\[
\frac{3^x}{27^{y-1}} = 81
\]

### Solving the First Equation:
The first equation is:

\[
\frac{5^x}{25^{3y-2}} = 1
\]

We know that \(25 = 5^2\), so rewrite the denominator:

\[
\frac{5^x}{(5^2)^{3y-2}} = \frac{5^x}{5^{2(3y-2)}} = \frac{5^x}{5^{6y-4}}
\]

Using the laws of exponents, simplify:

\[
5^{x - (6y - 4)} = 1
\]

Since \(5^0 = 1\), we can equate the exponent to zero:

\[
x - (6y - 4) = 0
\]

This simplifies to:

\[
x = 6y - 4 \quad \text{(Equation 1)}
\]

### Solving the Second Equation:
The second equation is:

\[
\frac{3^x}{27^{y-1}} = 81
\]

We know that \(27 = 3^3\) and \(81 = 3^4\), so rewrite the denominator and the right-hand side:

\[
\frac{3^x}{(3^3)^{y-1}} = 3^4
\]
\[
\frac{3^x}{3^{3(y-1)}} = 3^4
\]
\[
3^{x - 3(y-1)} = 3^4
\]

Since the bases are the same, equate the exponents:

\[
x - 3(y - 1) = 4
\]
\[
x - 3y + 3 = 4
\]
\[
x = 3y + 1 \quad \text{(Equation 2)}
\]

### Solving the System of Equations:
Now we have two equations:

1. \(x = 6y - 4\)
2. \(x = 3y + 1\)

Set them equal to each other:

\[
6y - 4 = 3y + 1
\]

Solve for \(y\):

\[
6y - 3y = 1 + 4
\]
\[
3y = 5
\]
\[
y = \frac{5}{3}
\]

Substitute \(y = \frac{5}{3}\) into one of the original equations, say \(x = 3y + 1\):

\[
x = 3\left(\frac{5}{3}\right) + 1 = 5 + 1 = 6
\]

Thus, the solution is:

\[
x = 6, \quad y = \frac{5}{3}
\]

Would you like more details on any part of the solution, or do you have any questions?

Here are five related questions to explore:
1. How do you solve simultaneous logarithmic equations?
2. What are the laws of exponents used to simplify expressions?
3. How do you solve equations involving fractional exponents?
4. What techniques are used to solve non-linear systems of equations?
5. How does changing the base of logarithms affect the solution?

**Tip:** When solving equations with exponents, always check if you can express the terms with a common base to simplify the process.

Solve the following simultaneous equations: 
 5^x / 25^(3y-2) = 1 
 3^x / 27^(y-1) = 81

This lesson provides a step-by-step solution to a simultaneous equation problem involving exponential expressions like 5^x and 3^x. The solution covers the laws of exponents and simplifying powers. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation