expand the logarithm as a sum or difference and or contant multiple of logarithms
log _4 16xy^2
thank you.. I know you do this:
log _4 16+ log _4 (x)+ log _4 (y^2)
and I know the first part = 2 and the last part you use the exponent rule =y^2log4
but the middle i don't get..thanks
For the last part of log4 (y^2)
you follow the pattern of
loga (x^n)
= n loga x
so log4 (y^2)
= 2 log4 y
the first part of your solution is ok
final answer:
2 + log4 x + 2 log4 y
love you guys
To expand the logarithm log₄(16xy²) as a sum or difference and/or constant multiple of logarithms, you can use the properties of logarithms.
First, we can simplify the expression log₄(16xy²).
We know that 16 can be written as 4², so we can rewrite the expression as log₄(4²xy²).
Using the exponent rule of logarithms, we can bring down the exponent 2 in front of the logarithm: log₄(4²) + log₄(xy²).
Since log₄(4²) = 2, the expression simplifies to 2 + log₄(xy²).
Now, let's focus on expanding log₄(xy²).
Since we have a product of xy², we can split it into two logarithms using the product rule: log₄(x) + log₄(y²).
Finally, we can write the expanded form of log₄(16xy²) as:
2 + log₄(x) + log₄(y²).
So, the middle part of the expansion is log₄(x).
To expand the logarithm log_4(16xy^2) as a sum or difference and/or constant multiple of logarithms, we can use the properties of logarithms.
1. Start by using the logarithm property log_a(b) + log_a(c) = log_a(b * c).
Therefore, log_4(16xy^2) = log_4(16) + log_4(x) + log_4(y^2).
2. Simplify the first term: log_4(16). Since 4^2 = 16, log_4(16) = 2.
3. The second term, log_4(x), does not have any further simplification.
4. The third term, log_4(y^2), can be further simplified using the logarithm property log_a(b^c) = c * log_a(b).
Therefore, log_4(y^2) = 2 * log_4(y).
Putting it all together, we have:
log_4(16xy^2) = 2 + log_4(x) + 2 * log_4(y).
Thus, the expanded form of log_4(16xy^2) as a sum or difference and/or constant multiple of logarithms is:
2 + log_4(x) + 2 * log_4(y).