Solving the Logarithmic Equation
We are asked to find the number of solutions for the equation \(\log_4(x-1) = \log_2(x - 3)\). To solve this, we need to work with logarithms that have the same base.
Applying Change of Base Formula
We can use the change of base formula for logarithms, which states that \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\). Let's change the base of \(\log_4(x-1)\) to base 2.
We know that \(4 = 2^2\). Using the formula:
\(\log_4(x-1) = \frac{\log_2(x-1)}{\log_2(4)}\)Since \(\log_2(4) = 2\), the expression becomes:
\(\log_4(x-1) = \frac{\log_2(x-1)}{2} = \frac{1}{2}\log_2(x-1)\)Rewriting and Simplifying the Equation
Now, substitute this back into the original equation:
\(\frac{1}{2}\log_2(x-1) = \log_2(x - 3)\)Multiply both sides by 2 to eliminate the fraction:
\(\log_2(x-1) = 2\log_2(x - 3)\)Using the logarithm power rule, \(n\log_b(a) = \log_b(a^n)\), we can rewrite the right side:
\(\log_2(x-1) = \log_2((x - 3)^2)\)Solving for x
Since the logarithms on both sides have the same base, their arguments must be equal:
\(x-1 = (x - 3)^2\)Expand the right side:
\(x-1 = x^2 - 6x + 9\)Rearrange the terms to form a standard quadratic equation (\(ax^2 + bx + c = 0\)):
\(0 = x^2 - 6x - x + 9 + 1\) \(x^2 - 7x + 10 = 0\)Factor the quadratic equation:
\((x-2)(x-5) = 0\)This gives us two potential solutions: \(x=2\) or \(x=5\).
Checking Domain Restrictions
It's crucial to check if these potential solutions are valid within the domains of the original logarithmic functions.
- For \(\log_4(x-1)\) to be defined, the argument must be positive: \(x-1 > 0 \implies x > 1\).
- For \(\log_2(x-3)\) to be defined, the argument must be positive: \(x-3 > 0 \implies x > 3\).
Both conditions must be satisfied, so the domain for the equation is \(x > 3\).
Verifying the Solutions
Now, let's check our potential solutions against the domain \(x > 3\).
- Check \(x=2\): Is \(2 > 3\)? No. Therefore, \(x=2\) is an extraneous solution and must be discarded.
- Check \(x=5\): Is \(5 > 3\)? Yes. Let's substitute \(x=5\) into the original equation to confirm:
- Left Hand Side (LHS): \(\log_4(5-1) = \log_4(4) = 1\)
- Right Hand Side (RHS): \(\log_2(5-3) = \log_2(2) = 1\)
Conclusion
After checking the potential solutions against the domain restrictions, we found that only \(x=5\) is a valid solution. Therefore, there is only one solution to the equation \(\log_4(x-1) = \log_2(x - 3)\).
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 ?
- 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 ?
- What is the smallest positive \(x\) satisfying \(\log_{\sin x} \cos x + \log_{\cos x} \sin x = 2\) ?
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 ?
