91Ó°ÊÓ

Logarithmic functions are the inverses of exponential functions. If you have an exponential function of the form \(y = b^x\), then its logarithmic form is \(x = \log_b(y)\). Here, \(\log_b(y)\) means 'the logarithm of \(y\) with base \(b\)'. Logarithmic functions are useful when you need to solve for the exponent.
For example, if you have \(2^3 = 8\), the logarithmic form would be \(3 = \log_2(8)\).

This means that the function \(f(x) = \log_b(x)\) will give you the exponent \(x\) needed for \(b\) to produce \(y\). The natural logarithm, denoted as \(\ln\), is a special type of logarithm where the base is Euler's number \(e\) (approximately 2.718). Hence, \(\ln(x)\) means \(\log_e(x)\).

Understanding logarithmic identities helps simplify complex expressions. Some key identities include:

• Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
• Quotient Rule: \(\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\)
• Power Rule: \(\log_b(M^k) = k \log_b(M)\)

In the context of the exercise, we specifically used the Power Rule.
Recalling the step-by-step solution:
Step 1: \(\ln(\sqrt{a}) = \ln(a^{\frac{1}{2}})\)
Step 2: Use the Power Rule: \(\ln(a^{\frac{1}{2}}) = \frac{1}{2} \ln(a)\).
This way, you can express \(\ln(\sqrt{a})\) as a product: \(\frac{1}{2} \ln(a)\).
Understanding these basic identities can greatly assist in solving logarithmic problems effectively.

Square roots and logarithms often intersect when dealing with expressions that involve roots. The square root of a number is the value that, when multiplied by itself, gives the original number. For instance, \(\sqrt{9} = 3\) as \(3 \times 3 = 9\).
In logarithms, roots can be expressed using exponents, allowing logarithmic identities to simplify these expressions.
For example, if you have \(\ln(\sqrt{a})\), you can convert the square root to an exponent: \(a^{\frac{1}{2}}\).
Then, using the logarithmic Power Rule: \(\ln(a^{\frac{1}{2}}) = \frac{1}{2} \ln(a)\).
This is exactly how we solved the exercise. Such transformations are vital in algebra and calculus, making complex expressions easier to handle and understand. By remembering these properties, solving logarithmic problems involving roots becomes more straightforward.