All Exams Test series for 1 year @ ₹349 only
Question

Given that $132^{0.14} = x$, $132^{0.26} = y$ and $x^z = y^2$, then the value of z is close to:

The correct answer is
3.71

Exponent Problem Analysis

We are given the following information:

  • $132^{0.14} = x$
  • $132^{0.26} = y$
  • $x^z = y^2$

Our goal is to find the value of $z$.

Solving for z: Step-by-Step

We can solve this problem by substituting the expressions for $x$ and $y$ into the equation $x^z = y^2$.

Step 1: Substitute x and y

Replace $x$ with $132^{0.14}$ and $y$ with $132^{0.26}$ in the equation $x^z = y^2$:

$ (132^{0.14})^z = (132^{0.26})^2 $

Step 2: Apply the Power of a Power Rule

We use the exponent rule $(a^m)^n = a^{m \times n}$. Applying this rule to both sides of the equation:

$ 132^{(0.14 \times z)} = 132^{(0.26 \times 2)} $

Step 3: Simplify the Exponents

Calculate the product on the right side:

$ 0.26 \times 2 = 0.52 $

The equation now becomes:

$ 132^{0.14z} = 132^{0.52} $

Step 4: Equate the Exponents

Since the bases are the same (132), the exponents must be equal:

$ 0.14z = 0.52 $

Step 5: Calculate the Value of z

To find $z$, divide both sides by 0.14:

$ z = \frac{0.52}{0.14} $

Simplify the fraction:

$ z = \frac{52}{14} = \frac{26}{7} $

Now, let's convert the fraction to a decimal:

$ z \approx 3.7142857... $

Comparing with Options

The calculated value of $z$ is approximately 3.714. Let's compare this with the given options:

  • 1. 5.03
  • 2. 4.84
  • 3. 2.37
  • 4. 3.71

The value 3.71 is the closest approximation to our calculated value of $z$.

Was this answer helpful?

Important Questions from Logarithms

  1. For \(x \ge y > 1\), let \(\log_x\left(\frac{x}{y}\right) + \log_y\left(\frac{y}{x}\right) = k\), then the value of \(k\) can never be equal to

  2. If \(\log_ba = p\), \(\log_dc = 2p\) and \(\log_fe = 3p\), then what is \((ace)^{\frac{1}{p}}\) equal to ?

  3. What is the number of solutions of \(\log_4(x-1) = \log_2(x - 3)\)?
  4. Let \(p = \ln(x)\), \(q = \ln(x^3)\) and \(r = \ln(x^5)\), where \(x > 1\). Which of the following statements is/are correct?
    I. \(p, q\) and \(r\) are in AP.
    II. \(p, q\) and \(r\) can never be in GP.
    Select the answer using the code given below.
  5. If \(a^b = b^a\), then what is 

    \(\frac{a \times   (\frac{a}{b}) ^{ \frac{a}{b} } }{ (a)^{ \frac{a}{b} } }\) 

    equal to?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App