If \(x + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\) then x is equal to
1
Understanding the Logarithm Equation
The problem asks us to find the value of \(x\) that satisfies the given equation involving logarithms:
\[x + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\]
To solve this equation, we need to use the properties of logarithms to simplify it and isolate the variable \(x\).
Applying Logarithm Properties
Recall the following logarithm properties:
- \(a \log_b c = \log_b c^a\)
- \(\log_b m + \log_b n = \log_b (mn)\)
- \(c = \log_b b^c\)
- If \(\log_b m = \log_b n\), then \(m = n\)
Let's rewrite the terms in the given equation using these properties, particularly converting terms without logs into log form with base 10, since all other log terms are base 10.
Rewrite \(x\) as \({\log _{10}}{10^x}\):
\[{\log _{10}}{10^x} + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\]
Rewrite \(x{\log _{10}}5\) as \({\log _{10}}{5^x}\):
\[{\log _{10}}{10^x} + {\log _{10}}\left( {1 + {2^x}} \right) = {\log _{10}}{5^x} + {\log _{10}}6\]
Now, apply the sum property (\(\log_b m + \log_b n = \log_b (mn)\)) to both sides of the equation:
Left side: \({\log _{10}}{10^x} + {\log _{10}}\left( {1 + {2^x}} \right) = {\log _{10}}\left( {10^x \cdot \left( {1 + {2^x}} \right)} \right)\)
Right side: \({\log _{10}}{5^x} + {\log _{10}}6 = {\log _{10}}\left( {{5^x} \cdot 6} \right)\)
So, the equation becomes:
\[{\log _{10}}\left( {10^x \left( {1 + {2^x}} \right)} \right) = {\log _{10}}\left( {6 \cdot {5^x}} \right)\]
Solving the Exponential Equation
Since the logarithms on both sides have the same base and are equal, their arguments must be equal:
\[{10^x}\left( {1 + {2^x}} \right) = 6 \cdot {5^x}\]
Expand the left side:
\[{10^x} + {10^x} \cdot {2^x} = 6 \cdot {5^x}\]
Recall that \(10^x = (2 \cdot 5)^x = 2^x \cdot 5^x\):
\[{2^x} \cdot {5^x} + ({2^x} \cdot {5^x}) \cdot {2^x} = 6 \cdot {5^x}\]
\[{2^x} \cdot {5^x} + {2^x} \cdot {2^x} \cdot {5^x} = 6 \cdot {5^x}\]
\[{2^x} \cdot {5^x} + {(2^x)^2} \cdot {5^x} = 6 \cdot {5^x}\]
Assuming \(5^x \neq 0\) (which is true for any real \(x\)), we can divide both sides by \(5^x\):
\[{2^x} + {(2^x)^2} = 6\]
Let \(y = 2^x\). The equation becomes a quadratic equation in terms of \(y\):
\[y + y^2 = 6\]
Rearrange it into the standard quadratic form \(ay^2 + by + c = 0\):
\[y^2 + y - 6 = 0\]
Factor the quadratic equation:
\[(y + 3)(y - 2) = 0\]
This gives two possible solutions for \(y\):
\[y + 3 = 0 \implies y = -3\]
\[y - 2 = 0 \implies y = 2\]
Now substitute back \(y = 2^x\):
Case 1: \(2^x = -3\). Since the base of the exponent (2) is positive, \(2^x\) must always be positive for any real value of \(x\). Therefore, \(2^x = -3\) has no real solution.
Case 2: \(2^x = 2\). This equation can be written as \(2^x = 2^1\). Since the bases are the same, the exponents must be equal:
\[x = 1\]
Thus, the only real solution to the given logarithm equation is \(x = 1\).
Verifying the Solution
Let's substitute \(x=1\) back into the original equation to verify:
\[x + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\]
Substitute \(x=1\):
\[1 + {\log _{10}}\left( {1 + {2^1}} \right) = 1 \cdot {\log _{10}}5 + {\log _{10}}6\]
\[1 + {\log _{10}}\left( {1 + 2} \right) = {\log _{10}}5 + {\log _{10}}6\]
\[1 + {\log _{10}}3 = {\log _{10}}5 + {\log _{10}}6\]
Using the property \(1 = \log_{10} 10\) on the left side:
\[{\log _{10}}10 + {\log _{10}}3 = {\log _{10}}5 + {\log _{10}}6\]
Using the sum property on both sides:
\[{\log _{10}}(10 \cdot 3) = {\log _{10}}(5 \cdot 6)\]
\[{\log _{10}}30 = {\log _{10}}30\]
The equation holds true for \(x=1\). Therefore, \(x=1\) is the correct solution.
figure class="table">| Step | Action | Equation |
|---|---|---|
| 1 | Start with the given equation. | \(x + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\) |
| 2 | Rewrite \(x\) as \({\log _{10}}{10^x}\). | \({\log _{10}}{10^x} + {\log _{10}}\left( {1 + {2^x}} \right) = x{\log _{10}}5 + {\log _{10}}6\) |
| 3 | Rewrite \(x{\log _{10}}5\) as \({\log _{10}}{5^x}\). | \({\log _{10}}{10^x} + {\log _{10}}\left( {1 + {2^x}} \right) = {\log _{10}}{5^x} + {\log _{10}}6\) |
| 4 | Apply sum property of logarithms. | \({\log _{10}}\left( {10^x (1 + {2^x})} \right) = {\log _{10}}\left( {6 \cdot {5^x}} \right)\) |
| 5 | Equate arguments of logarithms. | \({10^x} (1 + {2^x}) = 6 \cdot {5^x}\) |
| 6 | Expand and substitute \(10^x = 2^x \cdot 5^x\). | \({2^x} \cdot {5^x} + {(2^x)^2} \cdot {5^x} = 6 \cdot {5^x}\) |
| 7 | Divide by \(5^x\) (assuming \(5^x \neq 0\)). | \({2^x} + {(2^x)^2} = 6\) |
| 8 | Let \(y=2^x\) and rearrange into quadratic. | \(y^2 + y - 6 = 0\) |
| 9 | Factor the quadratic equation. | \((y+3)(y-2) = 0\) |
| 10 | Solve for \(y\). | \(y = -3\) or \(y = 2\) |
| 11 | Substitute back \(y=2^x\) and solve for \(x\). | \(2^x = -3\) (no real solution), \(2^x = 2 \implies x = 1\) |
The final answer is \(x=1\).
Revision Table: Key Logarithm Concepts
figure class="table">| Concept | Description | Example |
|---|---|---|
| Logarithm Definition | If \(b^y = x\), then \(\log_b x = y\). \(b > 0\), \(b \neq 1\), \(x > 0\). | \(10^2 = 100 \implies \log_{10} 100 = 2\) |
| Product Rule | \(\log_b (mn) = \log_b m + \log_b n\) | \(\log_{10} (5 \times 6) = \log_{10} 5 + \log_{10} 6\) |
| Power Rule | \(\log_b m^a = a \log_b m\) | \(x \log_{10} 5 = \log_{10} 5^x\) |
| Base Identity | \(\log_b b = 1\) | \(\log_{10} 10 = 1\) |
| Equating Logarithms | If \(\log_b m = \log_b n\), then \(m = n\) (where \(m, n > 0\)) | If \(\log_{10} A = \log_{10} B\), then \(A = B\) |
Additional Information: Solving Logarithmic Equations
Solving logarithmic equations often involves transforming the equation using logarithm properties to eliminate the logarithms. Common strategies include:
- Combining logarithmic terms on one side using the product or quotient rules.
- Moving terms without logarithms to the other side.
- Converting terms without logarithms into logarithmic form using the base identity (\(c = \log_b b^c\)).
- Using the power rule to move exponents.
- Once the equation is in the form \(\log_b M = \log_b N\) or \(\log_b M = c\), equate the arguments (\(M=N\)) or convert to exponential form (\(b^c = M\)), respectively.
- Solve the resulting algebraic or exponential equation.
- Always check your solutions in the original equation to ensure that the arguments of the logarithms are positive, as logarithms are only defined for positive numbers. In our case, the arguments were \(1 + 2^x\) and \(6 \cdot 5^x\). For real \(x\), \(2^x > 0\), so \(1+2^x > 1 > 0\). Also \(5^x > 0\), so \(6 \cdot 5^x > 0\). Thus, the arguments are always positive for real \(x\), and our solution \(x=1\) is valid.
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