If log 10 2 log 2 10 + log 10 (10 x) = 2, then what is the value of x?
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Solving Logarithmic Equations for the Value of x
The question asks us to find the value of \(x\) that satisfies the given logarithmic equation. Solving logarithmic equations often involves using various properties of logarithms to simplify the equation and isolate the variable.
The given equation is:
\(\log_{10} 2 \log_2 10 + \log_{10} (10 x) = 2\)
Key Logarithm Properties Used
To solve this equation, we will use the following important logarithm properties:
- Change of Base Formula: \(\log_b a = \frac{\log_c a}{\log_c b}\). A useful application is \(\log_b a = \frac{1}{\log_a b}\).
- Product Rule: \(\log_b (M N) = \log_b M + \log_b N\).
- Logarithm Definition: \(\log_b a = c\) is equivalent to \(b^c = a\).
- Identity Property: \(\log_b b = 1\).
Step-by-Step Solution to Find x
Let's solve the logarithmic equation step by step:
Step 1: Simplify the first term.
The first term is \(\log_{10} 2 \log_2 10\). Using the change of base property \(\log_b a = \frac{1}{\log_a b}\), we can write \(\log_2 10 = \frac{1}{\log_{10} 2}\).
So, the first term becomes:
\(\log_{10} 2 \times \frac{1}{\log_{10} 2}\)
Assuming \(\log_{10} 2\) is not zero (which it isn't, since \(10^0=1 \neq 2\)), we can cancel out \(\log_{10} 2\):
\(\log_{10} 2 \log_2 10 = 1\)
Step 2: Simplify the second term.
The second term is \(\log_{10} (10 x)\). Using the product rule \(\log_b (M N) = \log_b M + \log_b N\), we can write:
\(\log_{10} (10 x) = \log_{10} 10 + \log_{10} x\)
Using the identity property \(\log_b b = 1\), we know that \(\log_{10} 10 = 1\).
So, the second term simplifies to:
\(\log_{10} (10 x) = 1 + \log_{10} x\)
Step 3: Substitute the simplified terms back into the original equation.
The original equation is \(\log_{10} 2 \log_2 10 + \log_{10} (10 x) = 2\).
Substituting the simplified terms (1 for the first term and \(1 + \log_{10} x\) for the second term), we get:
\(1 + (1 + \log_{10} x) = 2\)
Step 4: Solve the resulting equation for \(\log_{10} x\).
\(1 + 1 + \log_{10} x = 2\)
\(2 + \log_{10} x = 2\)
Subtract 2 from both sides:
\(\log_{10} x = 2 - 2\)
\(\log_{10} x = 0\)
Step 5: Find the value of x using the definition of logarithm.
We have the equation \(\log_{10} x = 0\). Using the definition \(\log_b a = c \iff b^c = a\), where \(b=10\), \(a=x\), and \(c=0\), we can rewrite this as:
\(10^0 = x\)
Any non-zero number raised to the power of 0 is 1. So,
\(x = 1\)
Thus, the value of \(x\) that satisfies the equation is 1.
Comparing this result with the given options, we find that \(x=1\) corresponds to one of the choices.
Logarithm Properties Revision Table
| Property | Formula | Example |
|---|---|---|
| Change of Base | \(\log_b a = \frac{\log_c a}{\log_c b}\) | \(\log_2 8 = \frac{\log_{10} 8}{\log_{10} 2}\) |
| Inverse Property | \(\log_b a = \frac{1}{\log_a b}\) | \(\log_2 10 = \frac{1}{\log_{10} 2}\) |
| Product Rule | \(\log_b (MN) = \log_b M + \log_b N\) | \(\log_{10} (10x) = \log_{10} 10 + \log_{10} x\) |
| Quotient Rule | \(\log_b (\frac{M}{N}) = \log_b M - \log_b N\) | \(\log_2 (\frac{8}{4}) = \log_2 8 - \log_2 4\) |
| Power Rule | \(\log_b (M^k) = k \log_b M\) | \(\log_{10} (x^2) = 2 \log_{10} x\) |
| Identity Property | \(\log_b b = 1\) | \(\log_{10} 10 = 1\) |
| Zero Property | \(\log_b 1 = 0\) | \(\log_{10} 1 = 0\) |
| Definition | \(\log_b a = c \iff b^c = a\) | \(\log_{10} 100 = 2 \iff 10^2 = 100\) |
Additional Information on Logarithms and Solving Equations
- Logarithms are the inverse operation to exponentiation. The expression \(\log_b a\) answers the question "To what power must \(b\) be raised to get \(a\)?".
- The base \(b\) of a logarithm must be a positive number other than 1 (\(b > 0, b \neq 1\)). The argument \(a\) must be positive (\(a > 0\)).
- When solving logarithmic equations, it's crucial to check the domain. The arguments of all logarithms must be positive for the solution to be valid. In this problem, the original equation has \(\log_{10} 2\), \(\log_2 10\), and \(\log_{10} (10x)\). For \(\log_{10} (10x)\) to be defined, \(10x > 0\), which means \(x > 0\). Our solution \(x=1\) satisfies this condition (\(1 > 0\)), so it is a valid solution.
- Logarithms with base 10 are called common logarithms and are often written as \(\log x\) or \(\log_{10} x\). Logarithms with base \(e\) (Euler's number) are called natural logarithms and are written as \(\ln x\).
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