Given that $132^{0.14} = x$, $132^{0.26} = y$ and $x^z = y^2$, then the value of z is close to:
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 $z$.
Important Questions from Logarithms
- If \(p + q = 15\), then what is \(q-p\) equal to?
- If \(p + q = 66\), then which one of the following is correct?
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
If \(\log_ba = p\), \(\log_dc = 2p\) and \(\log_fe = 3p\), then what is \((ace)^{\frac{1}{p}}\) equal to ?
- What is the number of solutions of \(\log_4(x-1) = \log_2(x - 3)\)?
