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Question

Given that $50^{0.92} = x$, $50^{0.26}= y$ and $x^z = y^3$, then the value of z is close to:

The correct answer is
0.85

Understanding the Exponential Problem

We are given a problem involving exponents and variables. We need to find the value of $z$ based on the relationships provided:

  • $50^{0.92} = x$
  • $50^{0.26}= y$
  • $x^z = y^3$

The goal is to use the properties of exponents to solve for $z$.

Step-by-Step Solution

Follow these steps to find the value of $z$:

  1. Substitute x and y: Start with the equation $x^z = y^3$. Substitute the given expressions for $x$ and $y$ into this equation.
    Substitute $x = 50^{0.92}$ and $y = 50^{0.26}$: $ (50^{0.92})^z = (50^{0.26})^3 $
  2. Apply the Power of a Power Rule: Use the exponent rule $(a^m)^n = a^{m \times n}$ to simplify both sides of the equation.
    • Left side: $(50^{0.92})^z = 50^{0.92 \times z}$
    • Right side: $(50^{0.26})^3 = 50^{0.26 \times 3}$
    The equation now becomes: $ 50^{0.92z} = 50^{0.26 \times 3} $
  3. Simplify the Exponent on the Right Side: Calculate the product $0.26 \times 3$. $ 0.26 \times 3 = 0.78 $ So the equation is: $ 50^{0.92z} = 50^{0.78} $
  4. Equate the Exponents: Since the bases are the same (both are 50), the exponents must be equal. $ 0.92z = 0.78 $
  5. Solve for z: Isolate $z$ by dividing both sides by 0.92. $ z = \frac{0.78}{0.92} $

Final Calculation

Now, perform the division to find the value of $z$:

$ z = \frac{0.78}{0.92} $

To calculate this value:

$ z \approx 0.847826... $

Rounding this value to two decimal places, we get $z$ $\approx$ 0.85.

Therefore, the value of $z$ is close to 0.85.

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Important Questions from Surds and Indices

  1. Find the value of $47^{-8} \div 47^3 \times 47^{-19}$.

  2. Given that $58^{0.5} = x$, $58^{0.79}= y$ and $x^z = y^{10}$, then the value of z is close to:

  3. The value of $8^{2} +   \sqrt{ 3^{2} } -2 \sqrt{36} -11$ is

  4. Given that $81^{0.65} = x$, $81^{0.26}= y$ and $x^z = y^7$, then the value of z is close to:
  5. Find the value of $24^{-10} \div 24^{16} \times 24^{-7}$.
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