<h2>
The required "option A)
" is correct.</h2>
Step-by-step explanation:
We have,
![\log _{a}x](https://tex.z-dn.net/?f=%5Clog%20_%7Ba%7Dx)
To find, the value of
= ?
∴
, where a, b and x are positive
a ≠ 1 and b ≠ 1
We know that,
The logarithm identity,
![\log_{p}m=\dfrac{\log_{y}m}{\log_{y}p}](https://tex.z-dn.net/?f=%5Clog_%7Bp%7Dm%3D%5Cdfrac%7B%5Clog_%7By%7Dm%7D%7B%5Clog_%7By%7Dp%7D)
∴
= ![\dfrac{\log_{b}x}{\log_{b}a}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Clog_%7Bb%7Dx%7D%7B%5Clog_%7Bb%7Da%7D)
Where, b is the common base of logarithm
∴ The value of
= ![\dfrac{\log_{b}x}{\log_{b}a}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Clog_%7Bb%7Dx%7D%7B%5Clog_%7Bb%7Da%7D)
Thus, the required option A)
is correct.
Do the opposite of PEMDAS. First subtract b from both sides.
T = b + v²/2d
T (-b) = b (-b) + v²/2d
T - b = v²/2d
Multiply 2d to both sides
2d(T - b) = v²/2d(2d)
2d(T - b) = v²
Distribute 2d to both T and -b
2dT - 2db = v²
Isolate the v, root both sides
√(2dT - 2db) = √v²
v = √(2dT - 2db)
hope this helps
Answer:
x=340
y=-165
Step-by-step explanation:
x/10+y/5=1 Equation 1
x/8+y/6=15 Equation 2
Using elimination method, we multiply equation 1 by 1/6, the coefficient of y in equation 2 and then also multiply equation 2 by 1/5, the coefficient of y in equation 1, this will make you have the same coefficient and then easy to eliminate.
1/6(x/10+y/5=1)
x/60+y/30=1/6 Equation 3
1/5(x/8+y/6=15)
x/40+y/30=15/5
x/40+y/30=3 Equation 4
We subtract equation 3 from equation 4 to eliminate y
x/40-x/60= (3x-2x)/120= x/120
y/30-y/30
3-1/6= (18-1)/6= 17/6
We now have
x/120=17/6
Cross multiply
6x=2040
Divide both sides by 6
6x/6=2040/6
x=2040/6
x=340
Substitute x for 340 in equation 1
x/10+y/5=1
340/10+y/5=1
34+y/5=1
y/5=1-34
y/5=-33
Cross multiply
y=-33(5)
y=-165