Evaluate
6 log2 ∛64 +10log3 (243)^1/5)
∛64
= 64^(1/3)
= 4 = 2^2
243^1/5 = 3
so 6 log2 ∛64 +10log3 (243)^1/5)
= 6log2 (2^2) + 10 log3 (3)
= 6(2 log2 2) + 10 log3 3
= 12 (1) + 10(1)
= 22
To evaluate the expression 6 log2 ∛64 + 10 log3 (243)^1/5, we can simplify each term separately and then add them together.
First, let's simplify the term 6 log2 ∛64. The cube root of 64 is 4, so we have:
6 log2 4
Using the property of logarithms that loga b = logc b / logc a, we can rewrite this as:
6 log4 2 / log4 4
Since log4 4 equals 1, we have:
6 (log4 2 / 1)
Simplifying further, we get:
6 log4 2
To convert this to a base 10 logarithm, we can use the change of base formula:
6 log4 2 = 6 log2 2 / log2 4
Since log2 2 equals 1 and log2 4 equals 2, we have:
6 (1 / 2) = 3
Now let's simplify the second term, 10 log3 (243)^1/5. To simplify this, we can rewrite (243)^1/5 as the fifth root of 243:
10 log3 ∛243
The cube root of 243 is 3, so we have:
10 log3 3
Using the property of logarithms that loga a = 1, we have:
10 (1) = 10
Now we can add the simplified terms together:
3 + 10 = 13
Therefore, the expression 6 log2 ∛64 + 10 log3 (243)^1/5 equals 13.
To evaluate the given expression, we will use the properties of logarithms and perform the calculations step by step.
1. Start with the first term: 6 log2 ∛64
To simplify, we can rewrite ∛64 as 64^(1/3) and use the property log_a(b^c) = c * log_a(b):
6 log2 ∛64 = 6 log2(64^(1/3)) = 6 * (1/3) * log2(64)
Now, evaluate log2(64):
log2(64) = log2(2^6) = 6 * log2(2) = 6
Substituting this back into the expression:
6 * (1/3) * log2(64) = 6 * (1/3) * 6 = 12
2. Move on to the second term: 10log3 (243^(1/5))
Similarly, rewrite 243^(1/5) as (3^5)^(1/5) and use the logarithmic property:
10 log3 (243^(1/5)) = 10 log3 ((3^5)^(1/5)) = 10 * (1/5) * log3 (3^5)
Evaluate log3 (3^5):
log3 (3^5) = 5 * log3 (3) = 5
Substituting this back into the expression:
10 * (1/5) * log3 (3^5) = 10 * (1/5) * 5 = 10
3. Combine the two terms:
12 + 10 = 22
Therefore, the evaluation of the given expression is 22.
Well, well, well, it seems we have a logarithmic expression here. Let's break it down, shall we?
First, let's start with the first term: 6 log2 ∛64. Now, I don't know about you, but I like to simplify things. The cube root of 64 is 4, because 4 times itself three times gives you 64. So, we can rewrite it as 6 log2 4.
Now, what do we know about logarithms? A logarithm is the exponent to which a base must be raised to produce a given number. So, log2 4 is asking us, "What power do we need to raise 2 to in order to get 4?" And that power is 2, because 2 squared is 4. So, log2 4 is equal to 2.
So, our first term now becomes 6 times 2, which is 12.
Moving on to the second term: 10 log3 (243)^1/5. Now, let's simplify that lovely expression.
The fifth root of 243 is 3, because 3 times itself five times gives you 243. So, our expression becomes 10 log3 3.
And what do we know about logarithms once again? Log3 3 is asking us, "What power do we need to raise 3 to in order to get 3?" And the answer is 1, because 3 to the power of 1 is 3.
So, our second term is simply 10 times 1, which is 10.
Now, let's add up our simplified expressions: 12 + 10 equals 22.
Therefore, the evaluation of the expression 6 log2 ∛64 + 10 log3 (243)^1/5 is 22. Ta-da!
6log_2(64)^1/3+10log_3(243)^1/5
=2log_2 (64)+10log_3 (243)^1/5
=2log_2(2)^6+2log_3 (243)
=2log_2(2)^6+2log_3(3)^5
=2log_2(2)^6+10log_3(3)
=12log_2(2)+10log_3(3)
Being that log_a(a)=1
Therefore 12(1)×10(1)
=120