Log243-log64-log3 with the base 3,2,root3 give me a simplify of these
Question: log3 243 - log2 64 + log√3 3
This can also be written as:
log3 3^5 - log2 2^6 + log√3 √3^2
= 5log3 3 - 6log2 2 + 2log√3 √3 [Since loge m^n = nloge m]
= 5(1) - 6(1) + 2(1) [Since logm m = 1]
= 5-6+2
= 1
To simplify the expression log243-log64-log3 with different bases (base 3, base 2, and base √3), we need to use the rules of logarithms.
First, let's simplify each term separately using the given bases:
1. log243 with base 3:
To find the exponent to which 3 must be raised to obtain 243, we get:
3^x = 243
x = log3(243)
Since 3^5 = 243, we have:
log243 with base 3 = 5
2. log64 with base 2:
To find the exponent to which 2 must be raised to obtain 64, we get:
2^y = 64
y = log2(64)
Since 2^6 = 64, we have:
log64 with base 2 = 6
3. log3 with base √3:
To find the exponent to which √3 must be raised to obtain 3, we get:
√3^z = 3
z = log√3(3)
Since (√3)^2 = 3, we have:
log3 with base √3 = 2
Now, let's substitute the simplified values back into the original expression:
log243-log64-log3 = 5-6-2
To evaluate this expression, we simply subtract the values:
5 - 6 - 2 = -3
Therefore, the simplified expression of log243-log64-log3 with the bases 3, 2, and √3 is -3.
log 3 9 +log 243 9+2log 3 9
I read that as
log3 243 - log2 64 - log√3 3
notice that 3^5 = 243 , 2^6 = 64 and (√3)^2 = 3
then:
log3 243 - log2 64 - log√3 3
= 5 - 6 - 2
= -3