Solve Log₃11/Log₃4 + 2log3/ln3: Logarithmic Expression Challenge

Question

log311log34+1ln32log3= \frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3=

Video Solution

Solution Steps

00:00 Solve
00:03 Let's use the formula for logarithmic division
00:07 We'll get the logarithm of the numerator in base of the denominator
00:13 We'll convert from ln to log
00:20 When dividing 1 by log, we get the inverse log in number and base
00:35 We'll use the formula for logarithmic multiplication, switching between bases
00:50 The logarithm of a number in its own base is always equal to 1
01:01 We'll use the formula for logarithm of a power, putting 2 inside the log
01:10 And this is the solution to the question

Step-by-Step Solution

To solve this problem, we'll proceed as follows:

  • Step 1: Rewrite each logarithmic expression using the change of base formula.
  • Step 2: Simplify the expressions using properties of logarithms.
  • Step 3: Identify the final expression.

Now, let's work through each step:

Step 1: We begin by converting each logarithm to the natural logarithm base.
Using the change of base formula, we have:

log311log34=ln11ln3ln4ln3=ln11ln4 \frac{\log_3 11}{\log_3 4} = \frac{\frac{\ln 11}{\ln 3}}{\frac{\ln 4}{\ln 3}} = \frac{\ln 11}{\ln 4}.

Step 2: Next, simplify the second expression:

1ln32log3=2 \frac{1}{\ln 3} \cdot 2\log 3 = 2.

This follows because log3\log 3 in natural logarithms converts to ln3\ln 3, and thus:

2ln3ln3=2 \frac{2\ln 3}{\ln 3} = 2.

Hence, our entire expression now is ln11ln4+2\frac{\ln 11}{\ln 4} + 2.

Step 3: Express 22 as a logarithm. Using the properties of logarithms:

2=loge22 = \log e^2, since lne=1\ln e = 1.

Therefore, the entire expression becomes:

ln11ln4+loge2 \frac{\ln 11}{\ln 4} + \log e^2.

By the properties of logarithms, this can also be expressed as:

log411+loge2 \log_4 11 + \log e^2.

Thus, the expression simplifies directly to:

log411+loge2 \log_4 11 + \log e^2.

Therefore, the solution to the problem is log411+loge2 \log_4 11 + \log e^2 .

Answer

log411+loge2 \log_411+\log e^2


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