Question

Evaluate the logarithm. Round your result to three decimal places.\(\log _{27} 35\)

Short answer

1.280

Step-by-step solution

Understand the problem

Here we have to evaluate \(\log _{27}35\). This means we have to find the exponent that 27 must be raised to obtain 35.

Conversion to natural logarithm

It is easier to calculate logarithms with base \(e\), i.e., natural logarithms (\(\ln\)). Logarithms with any base can be converted to natural logarithms using the formula \(\log_a b = \frac{\ln b}{\ln a}\). So, \(\log _{27}35 = \frac{\ln 35}{\ln 27}\)

Calculate using calculator

Now you can calculate both natural logarithms \(\ln 35\) and \(\ln 27\), and then divide the former by the latter. Don’t forget to round the answer to three decimal places. Using a scientific calculator you get \(\frac{\ln 35}{\ln 27} ≈ 1.280\)

Key concepts

Logarithm Properties

Logarithms are like the 'detectives' of the math world – they help us find the missing exponent needed to get from one number to another. But they have a set of rules, or properties, that make working with them easier. Here are the crucial ones to remember:

• Product Rule: The logarithm of a product is equal to the sum of the logarithms of the factors. In math terms, this is \(\log_b(mn) = \log_b(m) + \log_b(n)\).
• Quotient Rule: The logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator. So, \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\).
• Power Rule: The logarithm of a power is equal to the exponent times the logarithm of the base, represented as \(\log_b(m^k) = k \cdot \log_b(m)\).
• Change of Base Formula: This is handy for converting logarithms from one base to another and is especially useful for our exercise. It's written as \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\), where 'c' can be any positive number. For calculators, it's common to use base 10 or the natural logarithm base, 'e'.

Understanding these properties isn't just about memorizing them; it's about seeing how they make sense. Just like you know that 2 apples plus 2 apples make 4 apples, these properties let you work with logarithms to simplify and solve problems.

Natural Logarithm

The natural logarithm, written as \(\ln\), is a special kind of logarithm because its base is 'e'. The number 'e' is an irrational number, just like pi, and it's about 2.718. Think of 'e' as a universal constant that pops up in growth processes, like interest in a bank or bacteria multiplying.

Why do we care about the natural logarithm? Because it makes complex calculations, especially those involving growth and decay, much simpler. Calculus loves it for its nice properties when dealing with derivatives and integrals. In terms of our problem, using the formula \(\log_a(b) = \frac{\ln(b)}{\ln(a)}\), we swap awkward base 27 for the friendly base 'e'. This isn't just to make life easy for calculators; it's actually a more natural way to express logarithmic relationships and is deeply connected to the way the world around us changes and grows.

Logarithmic Conversion

Sometimes, you'll run into a logarithm with a base that's not 10 or 'e', and you'll need to convert it. That's what logarithmic conversion is all about. Basically, it's a way to reframe the question to make it easier to solve with the tools we have – like a calculator that only has buttons for \(\log\) (base 10) and \(\ln\) (base e).

In the exercise \(\log _{27}35\), we used the change of base formula to switch from a base of 27 to 'e'. By doing so, we could punch the numbers into a calculator with an \(\ln\) function and get an answer.

• We took the original logarithm \(\log_{27}35\) and applied the change of base formula.
• We calculated the natural logs of 35 and 27 and put one over the other.
• We rounded our final answer to three decimal places because that's often enough precision for practical purposes, and also because sometimes the numbers just keep going with no end in sight (those pesky irrational numbers again!).

When you grasp logarithmic conversion, it's like getting a translator for a language you don't speak. It may seem like a detour, but it's really a bridge to finding the answer with the tools at hand.