We are asked to find the smallest positive \(x\) that satisfies the equation:
\( \log_{\sin x} \cos x + \log_{\cos x} \sin x = 2 \)Logarithm Equation Simplification
To solve this equation, we can use a substitution. Let \(y = \log_{\sin x} \cos x\). Using the change of base property for logarithms, \(\log_b a = \frac{1}{\log_a b}\), we have \(\log_{\cos x} \sin x = \frac{1}{\log_{\sin x} \cos x} = \frac{1}{y}\).
Substituting this into the original equation gives:
\( y + \frac{1}{y} = 2 \)Solve for Substitution Variable
To find the value of \(y\), we can solve this equation:
- Multiply both sides by \(y\) (assuming \(y \neq 0\)): \(y^2 + 1 = 2y\)
- Rearrange the terms to form a quadratic equation: \(y^2 - 2y + 1 = 0\)
- Factor the quadratic equation: \((y - 1)^2 = 0\)
- Solve for \(y\): \(y = 1\)
Solve for x Value
Now, substitute back \(y = \log_{\sin x} \cos x\):
\( \log_{\sin x} \cos x = 1 \)By the definition of a logarithm, if \(\log_b a = c\), then \(a = b^c\). Applying this here:
\( \cos x = (\sin x)^1 \) \( \cos x = \sin x \)To find \(x\), we can divide both sides by \(\cos x\) (assuming \(\cos x \neq 0\)):
\( \frac{\sin x}{\cos x} = 1 \) \( \tan x = 1 \)The general solution for \(\tan x = 1\) is \(x = n\pi + \frac{\pi}{4}\), where \(n\) is an integer.
We need the smallest positive value for \(x\). Setting \(n=0\) gives:
\( x = 0\pi + \frac{\pi}{4} = \frac{\pi}{4} \)Domain Constraints Check
For the original logarithmic equation \(\log_{\sin x} \cos x + \log_{\cos x} \sin x = 2\) to be defined, the bases (\(\sin x\), \(\cos x\)) must be positive and not equal to 1, and the arguments (\(\cos x\), \(\sin x\)) must be positive.
- \(\sin x > 0\) and \(\sin x \neq 1\)
- \(\cos x > 0\) and \(\cos x \neq 1\)
These conditions require \(x\) to be in the first quadrant, i.e., \(0 < x < \frac{\pi}{2}\).
Our potential solution is \(x = \frac{\pi}{4}\). Let's check if it satisfies the conditions:
- \(x = \frac{\pi}{4}\) is in the interval \((0, \frac{\pi}{2})\).
- \(\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\), which is positive and not equal to 1.
- \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\), which is positive and not equal to 1.
Since \(x = \frac{\pi}{4}\) satisfies the domain requirements and the equation \(\tan x = 1\), it is the smallest positive solution.
Similar Questions
- 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 \(\log_4(x-1) = \log_2(x - 3)\)?
- 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. - If \(1 - \log_{10}2 = \log_{10}(5^x + 4^x + 3^x + 2^x + 1)\), then what is a value of x ?
Important Questions from Logarithms
Given that $132^{0.14} = x$, $132^{0.26} = y$ and $x^z = y^2$, then the value of z is close to:
- 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 ?
