Math Problem Statement

Suppose 2^{x+y} = 125 and 2^{x-y} = 25, without using a calculator determine in closed form, the value of 2^{y/x}.

Solution

We are given the following two equations:

  1. 2x+y=1252^{x+y} = 125
  2. 2xy=252^{x-y} = 25

We are asked to determine, in closed form, the value of 2yx2^{\frac{y}{x}}.

Step 1: Express the equations in logarithmic form

We know that 125=53125 = 5^3 and 25=5225 = 5^2, so we can rewrite the equations as follows:

2x+y=53and2xy=522^{x+y} = 5^3 \quad \text{and} \quad 2^{x-y} = 5^2

Step 2: Take the logarithm on both sides

Taking the natural logarithm (or log base 2 for simplicity) on both sides of both equations, we get:

x+y=log2125andxy=log225x + y = \log_2 125 \quad \text{and} \quad x - y = \log_2 25

Step 3: Solve the system of equations

We now have the system of linear equations:

x+y=log2125(1)x + y = \log_2 125 \quad (1) xy=log225(2)x - y = \log_2 25 \quad (2)

Add equations (1) and (2) to eliminate yy:

(x+y)+(xy)=log2125+log225(x + y) + (x - y) = \log_2 125 + \log_2 25 2x=log2(125×25)=log231252x = \log_2 (125 \times 25) = \log_2 3125

Now, solve for xx:

x=12log23125x = \frac{1}{2} \log_2 3125

Since 3125=553125 = 5^5, we have:

x=12log2(55)=12×5log25=5log252x = \frac{1}{2} \log_2 (5^5) = \frac{1}{2} \times 5 \log_2 5 = \frac{5 \log_2 5}{2}

Step 4: Solve for yy

Now subtract equation (2) from equation (1) to eliminate xx:

(x+y)(xy)=log2125log225(x + y) - (x - y) = \log_2 125 - \log_2 25 2y=log2(12525)=log252y = \log_2 \left(\frac{125}{25}\right) = \log_2 5

So:

y=12log25y = \frac{1}{2} \log_2 5

Step 5: Find 2yx2^{\frac{y}{x}}

Now we need to compute 2yx2^{\frac{y}{x}}. Using the values we found for yy and xx:

yx=12log2552log25=15\frac{y}{x} = \frac{\frac{1}{2} \log_2 5}{\frac{5}{2} \log_2 5} = \frac{1}{5}

Thus:

2yx=2152^{\frac{y}{x}} = 2^{\frac{1}{5}}

Final Answer:

The value of 2yx2^{\frac{y}{x}} is 2152^{\frac{1}{5}}, which is the fifth root of 2.

Would you like any additional clarifications or details?

Here are some related questions to explore:

  1. How can we use logarithms to solve other types of exponential equations?
  2. What are the properties of logarithmic functions that help solve exponential systems?
  3. What other methods can be used to solve systems of nonlinear equations like this?
  4. How would the solution change if the exponents were fractions instead of integers?
  5. Can this method be applied to exponential equations involving different bases, like base 3?

Tip: Always check if numbers can be rewritten as powers of smaller primes (like 2, 3, or 5) to simplify exponential equations.

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Math Problem Analysis

Mathematical Concepts

Exponential Equations
Logarithms
Algebra

Formulas

x + y = log_2 125
x - y = log_2 25
2x = log_2 3125
2y = log_2 5

Theorems

Logarithm properties (log(a*b) = log(a) + log(b))
Basic exponent and logarithm relations

Suitable Grade Level

Grade 10-12 or early university level