Solving Logarithm Equation for Integer x

The problem asks us to find the integer value of $x$ that satisfies the given equation:

\(\qquad x + \log_{15} (1 + 3^x) = x \log_{15} 5 + \log_{15} 12\)

Step-by-Step Solution Process

We need to manipulate the equation using properties of logarithms to isolate $x$. The key properties we will use are:

• \(a \log_b c = \log_b c^a\)
• \(\log_b c + \log_b d = \log_b (c \cdot d)\)
• \(\log_b b = 1\), so \(a = a \cdot 1 = a \log_b b = \log_b b^a\)

Let's rewrite each term in the equation using a common base, which is 15 in this case.

Rewrite $x$ as a logarithm with base 15:

\(\qquad x = x \cdot 1 = x \cdot \log_{15} 15 = \log_{15} 15^x\)

Rewrite the term \(x \log_{15} 5\) using the power property:

\(\qquad x \log_{15} 5 = \log_{15} 5^x\)

Now, substitute these back into the original equation:

\(\qquad \log_{15} 15^x + \log_{15} (1 + 3^x) = \log_{15} 5^x + \log_{15} 12\)

Apply the sum property of logarithms (\(\log_b c + \log_b d = \log_b (c \cdot d)\)) to both sides of the equation:

Left side:

\(\qquad \log_{15} 15^x + \log_{15} (1 + 3^x) = \log_{15} [15^x (1 + 3^x)]\)

Right side:

\(\qquad \log_{15} 5^x + \log_{15} 12 = \log_{15} (5^x \cdot 12)\)

Equating the two sides, we get:

\(\qquad \log_{15} [15^x (1 + 3^x)] = \log_{15} (5^x \cdot 12)\)

Since the logarithms on both sides have the same base, their arguments must be equal:

\(\qquad 15^x (1 + 3^x) = 5^x \cdot 12\)

Expand the left side:

\(\qquad 15^x \cdot 1 + 15^x \cdot 3^x = 5^x \cdot 12\)

\(\qquad 15^x + (15 \cdot 3)^x = 12 \cdot 5^x\)

\(\qquad 15^x + 45^x = 12 \cdot 5^x\)

Now, divide the entire equation by \(5^x\) (since \(5^x\) is always positive for real $x$, we don't need to worry about dividing by zero):

\(\qquad \frac{15^x}{5^x} + \frac{45^x}{5^x} = \frac{12 \cdot 5^x}{5^x}\)

Using the property \((\frac{a}{b})^x = \frac{a^x}{b^x}\):

\(\qquad \left(\frac{15}{5}\right)^x + \left(\frac{45}{5}\right)^x = 12\)

\(\qquad 3^x + 9^x = 12\)

Notice that \(9^x = (3^2)^x = (3^x)^2\). Let \(y = 3^x\). The equation becomes:

\(\qquad y + y^2 = 12\)

Rearrange this into a standard quadratic equation form:

\(\qquad y^2 + y - 12 = 0\)

Now, solve the quadratic equation for $y$. We can factor the quadratic. We look for two numbers that multiply to -12 and add up to 1. These numbers are 4 and -3.

\(\qquad (y + 4)(y - 3) = 0\)

This gives two possible solutions for $y$:

• \(y + 4 = 0 \implies y = -4\)
• \(y - 3 = 0 \implies y = 3\)

Substitute back \(y = 3^x\) to find the values of $x$:

Case 1: \(3^x = -4\). An exponential function with a positive base (\(3^x\)) cannot produce a negative value. Therefore, this case yields no real solution for $x$.

Case 2: \(3^x = 3\). Since \(3 = 3^1\), we have \(3^x = 3^1\). Because the bases are equal, the exponents must be equal.

\(\qquad x = 1\)

The problem states that $x$ is an integer. The solution we found, $x=1$, is an integer. Let's quickly verify this solution in the original equation.

For $x=1$:

Left side: \(1 + \log_{15}(1 + 3^1) = 1 + \log_{15}(1 + 3) = 1 + \log_{15} 4\). We can write $1$ as \(\log_{15} 15\). So, the left side is \(\log_{15} 15 + \log_{15} 4 = \log_{15}(15 \cdot 4) = \log_{15} 60\).

Right side: \(1 \log_{15} 5 + \log_{15} 12 = \log_{15} 5 + \log_{15} 12 = \log_{15}(5 \cdot 12) = \log_{15} 60\).

Since the left side equals the right side, $x=1$ is indeed the correct solution.

The integer value of $x$ is 1.

Revision Table: Logarithm Properties Used

Additional Information: Solving Exponential Equations

The equation \(3^x + 9^x = 12\) is a type of exponential equation. By recognizing that \(9^x = (3^x)^2\), we transformed it into a quadratic equation in terms of \(3^x\). This is a common technique for solving exponential equations that involve related bases (like 3 and 9, where 9 is a power of 3).

When solving equations involving exponential terms like \(a^x\), remember that if $a > 0$, then \(a^x\) is always positive for any real value of $x$. This is why the solution \(3^x = -4\) was discarded, as \(3^x\) can never be negative.

Solving the quadratic equation \(y^2 + y - 12 = 0\) was essential. The methods include factoring, using the quadratic formula (\(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)), or completing the square. Factoring was the quickest method here.

The final answer is $x=1$.