Understanding the Arithmetic Progression Property

The question states that the three given terms, $\log_{10} 2$, $\log_{10} (2^x - 1)$, and $\log_{10} (2^x + 3)$, are in an Arithmetic Progression (AP). An important property of an AP is that if three terms $a, b, c$ are in AP, then the middle term $b$ is the average of the first and third terms, i.e., $b = \frac{a+c}{2}$. This can be rewritten as $2b = a + c$.

Using this property with the given terms:

• First term ($a$): $\log_{10} 2$
• Second term ($b$): $\log_{10} (2^x - 1)$
• Third term ($c$): $\log_{10} (2^x + 3)$

We apply the property $2b = a + c$:

$\qquad 2 \log_{10} (2^x - 1) = \log_{10} 2 + \log_{10} (2^x + 3)$

Applying Logarithm Properties to Simplify

To solve this equation, we need to use standard logarithm properties:

• Power Rule: $k \log_b M = \log_b M^k$
• Product Rule: $\log_b M + \log_b N = \log_b (MN)$
• Equality Property: If $\log_b M = \log_b N$, then $M = N$ (assuming $M, N > 0$)

Apply the Power Rule to the left side of our equation:

$\qquad \log_{10} (2^x - 1)^2 = \log_{10} 2 + \log_{10} (2^x + 3)$

Apply the Product Rule to the right side of the equation:

$\qquad \log_{10} (2^x - 1)^2 = \log_{10} [2 \cdot (2^x + 3)]$

Now, since the logarithms on both sides have the same base (base 10), we can use the Equality Property to equate their arguments:

$\qquad (2^x - 1)^2 = 2(2^x + 3)$

Solving the Resulting Equation for x

Let's expand and rearrange the equation:

$\qquad (2^x)^2 - 2 \cdot 2^x \cdot 1 + 1^2 = 2 \cdot 2^x + 2 \cdot 3$

$\qquad (2^x)^2 - 2 \cdot 2^x + 1 = 2 \cdot 2^x + 6$

This equation involves the term $2^x$. Let's make a substitution to simplify it. Let $y = 2^x$. The equation then becomes a quadratic equation in terms of $y$:

$\qquad y^2 - 2y + 1 = 2y + 6$

Move all terms to one side to get a standard quadratic form $ay^2 + by + c = 0$:

$\qquad y^2 - 2y - 2y + 1 - 6 = 0$

$\qquad y^2 - 4y - 5 = 0$

Now we solve this quadratic equation for $y$. We can factor it:

$\qquad (y - 5)(y + 1) = 0$

This gives two possible values for $y$:

• $y - 5 = 0 \implies y = 5$
• $y + 1 = 0 \implies y = -1$

Finding x and Checking for Valid Solutions

Now we substitute back $2^x$ for $y$ to find the value of $x$:

Case 1: $y = 5$

$\qquad 2^x = 5$

To solve for $x$, we can take the logarithm base 2 of both sides:

$\qquad \log_2 (2^x) = \log_2 5$

Using the property $\log_b b^k = k$, we get:

$\qquad x = \log_2 5$

Case 2: $y = -1$

$\qquad 2^x = -1$

The exponential function $2^x$ always yields a positive value for any real number $x$. Therefore, $2^x = -1$ has no real solution for $x$. This case does not yield a valid value for $x$ in the real number system.

Finally, we must check if the solution $x = \log_2 5$ makes the original logarithmic terms defined. The argument of a logarithm must be positive. The terms are $\log_{10} 2$, $\log_{10} (2^x - 1)$, and $\log_{10} (2^x + 3)$.

• The argument of the first term is $2$, which is positive.
• The argument of the second term is $2^x - 1$. For $x = \log_2 5$, $2^x = 2^{\log_2 5} = 5$. So, $2^x - 1 = 5 - 1 = 4$. Since $4 > 0$, this term is defined.
• The argument of the third term is $2^x + 3$. For $x = \log_2 5$, $2^x = 5$. So, $2^x + 3 = 5 + 3 = 8$. Since $8 > 0$, this term is defined.

Since all arguments are positive for $x = \log_2 5$, this is a valid solution.

Thus, $x$ is equal to $\log_2 5$.

Revision Table: Key Concepts for Solving the Problem

Additional Information: Understanding Logarithms and AP

Logarithms are the inverse operations to exponentiation. The expression $\log_b M$ asks "To what power must we raise the base $b$ to get $M$?". For $\log_{10} 2$, it's the power we raise 10 to get 2. For $\log_{10} (2^x - 1)$, it's the power we raise 10 to get $2^x - 1$.

An Arithmetic Progression is a sequence where the difference between consecutive terms is constant. For example, $2, 5, 8, 11, \dots$ is an AP with a common difference of 3. If terms $a, b, c$ are in AP, the common difference is $b-a = c-b$. Rearranging this gives $2b = a+c$. This fundamental property was key to setting up the equation in this problem.

The substitution $y = 2^x$ is a common technique used to transform an equation involving an exponential term into a simpler form, such as a quadratic equation, which we can then solve using standard algebraic methods.