91Ó°ÊÓ

Exponential equations are equations where variables appear as exponents or in the base of exponential expressions. Solving logarithmic equations often involves rewriting them in exponential form. This is because the logarithm is the inverse operation of exponentiation.
For example, in the problem we are given

• \(\log_{2}(x+25) = 4\)

To find the value of \(x\), we rewrite the logarithmic equation as an exponential equation. The base \(2\) raised to the power \(4\) should equal \(x+25\). Therefore, we have:

• \(x + 25 = 2^4\)

This means \(2^4 = 16\), giving us the equation \(x + 25 = 16\). By rearranging this equation, we can solve for \(x\) and find that \(x = -9\). This method of changing forms helps us understand how logarithms and exponential functions are inherently related.

The domain of logarithmic functions is essential as a logarithm is defined only for positive numbers. When solving logging equations, we must ensure that all values make the logarithmic expression valid.
In the problem given, the logarithm is \(\log_{2}(x+25)\). Therefore, the expression inside the logarithm, \(x+25\), must be greater than zero:

• \(x + 25 > 0\)

Solving this inequality shows that \(x > -25\). This means that any solution for \(x\) must be greater than \(-25\) to be within the domain of the logarithmic function. In our case, we found \(x = -9\), which satisfies this condition since

• \(-9 + 25 = 16\)

This is indeed greater than zero, confirming that our solution is valid within the domain.

Solution verification in logarithmic equations involves checking that the solution obtained is correct and valid within the context of the problem. Every solution must satisfy both the equation and the constraints imposed by the domain of the logarithm.
For our solution, when we solved

• \(x + 25 = 16\)

and found \(x = -9\), we needed to ensure it was within the valid range for the logarithm. Substituting back, \(-9 + 25 = 16\), we see that 16 is different from zero, satisfying the inequality constraint of the domain \(x + 25 > 0\). This step guarantees that our solution is both mathematically correct and consistent with the restrictions of logarithmic operations.

Decimal approximation is used when an exact numerical solution might be difficult to interpret, requiring the solution to be expressed in decimal form for clarity. However, this conversion can sometimes be redundant when the answer is already a clear whole number or easily expressed value.
In our solved equation, \(x\) was computed as

• \(-9\)

Since this is a whole number, the approximation simply becomes:

• \(-9.00\)

Always, if a problem asks for a solution up to two decimal places, it is key to add any necessary zeros after the decimal for completeness. This helps maintain consistency and practice if future problems involve non-integers.