12, 4
take the first number together than the last ones and just find the difference
Given:
4log1/2^w (2log1/2^u-3log1/2^v)
Req'd:
Single logarithm = ?
Sol'n:
First remove the parenthesis,
4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)
Simplify each term,
Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:
log 1/2 (w^4) + 2 log 1/2 (u) - 3 log 1/2 (v)
log 1/2 (w^4) + log 1/2 (u^2) - log 1/2 (v^3)
We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):
Thus,
Log of 1/2 (w^4 u^2) - log of 1/2 (v^3)
then use the quotient property of logarithms which is log of b (x) - log of b (y) = log of b (x/y)
Therefore,
log of 1/2 (w^4 u^2 / v^3)
and for the final step and answer, reorder or rearrange w^4 and u^2:
log of 1/2 (u^2 w^4 / v^3)
Answer:
d<-1.75
Step-by-step explanation:
The two numbers are (7+ sqrt(65))/2 and (7- sqrt(65))/2
(7+ sqrt(65))/2+ (7- sqrt(65))/2= 7
(7+ sqrt(65))/2* (7- sqrt(65))/2)= -4.
Hope this helps~
