Answer:
log3 81 = 4
Step-by-step explanation:
Convert the exponential equation to a logarithmic equation using the logarithm base (3)(3) of the right side (81)(81) equals the exponent (4)(4).
log3(81)=4
or
you can remember this
loga Y= X
so, a^x =y
Answer:
answer is 105 hope u like it
Answer:
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Step-by-step explanation:
Answer:
![\displaystyle \large \boxed{x = \frac{ - 3 \pm \sqrt{ 7}i }{2} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%20%5Cboxed%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%207%7Di%20%20%7D%7B2%7D%20%7D)
Step-by-step explanation:
We are given the equation:
![\displaystyle \large{ {x}^{2} + 3x + 4 = 0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%203x%20%2B%204%20%3D%200%7D)
Since the expression is not factorable with real numbers, we use the Quadratic Formula.
<u>Q</u><u>u</u><u>a</u><u>d</u><u>r</u><u>a</u><u>t</u><u>i</u><u>c</u><u> </u><u>F</u><u>o</u><u>r</u><u>m</u><u>u</u><u>l</u><u>a</u>
![\displaystyle \large{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%20b%20%5Cpm%20%20%5Csqrt%7B%20%7Bb%7D%5E%7B2%7D%20%20-%204ac%7D%20%7D%7B2a%7D%20%7D)
Compare the expression:
![\displaystyle \large{a {x}^{2} + bx + c = {x}^{2} + 3x + 4 }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7Ba%20%7Bx%7D%5E%7B2%7D%20%2B%20bx%20%2B%20c%20%3D%20%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%203x%20%2B%204%20%7D)
a = 1
b = 3
c = 4
Substitute a = 1, b = 3 and c = 4 in the formula.
![\displaystyle \large{x = \frac{ - 3 \pm \sqrt{ {3}^{2} - 4(1)(4)} }{2(1)} } \\ \displaystyle \large{x = \frac{ - 3 \pm \sqrt{ 9 - 16} }{2} } \\ \displaystyle \large{x = \frac{ - 3 \pm \sqrt{ - 7} }{2} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%20%7B3%7D%5E%7B2%7D%20%20-%204%281%29%284%29%7D%20%7D%7B2%281%29%7D%20%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%209%20%20-%2016%7D%20%7D%7B2%7D%20%7D%20%5C%5C%20%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%20%20-%207%7D%20%7D%7B2%7D%20%7D)
<u>I</u><u>m</u><u>a</u><u>g</u><u>i</u><u>n</u><u>a</u><u>r</u><u>y</u><u> </u><u>U</u><u>n</u><u>i</u><u>t</u>
![\displaystyle \large{i = \sqrt{ - 1} }](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Clarge%7Bi%20%3D%20%20%5Csqrt%7B%20-%201%7D%20%7D)
Therefore,
![\displaystyle \large{x = \frac{ - 3 \pm \sqrt{ 7} \sqrt{ - 1} }{2} } \\ \displaystyle \large{x = \frac{ - 3 \pm \sqrt{ 7}i }{2} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%207%7D%20%5Csqrt%7B%20-%201%7D%20%20%7D%7B2%7D%20%7D%20%5C%5C%20%5Cdisplaystyle%20%5Clarge%7Bx%20%3D%20%20%5Cfrac%7B%20-%203%20%5Cpm%20%20%5Csqrt%7B%207%7Di%20%20%7D%7B2%7D%20%7D)
here,
Q is the midpoint of PR.
O is the midpoint of PN.
So, the line joining the midpoints of two sides is half and parallel to the third side.
So, QO is parallel to NR.
Now,
in triangles OPQ and NPR,
i) angle OPQ = angle NPR (common angle)
ii) angle POQ = angle PNR (corresponding angles)
iii) angle OQP = angle NRP (corresponding angles)
so,
triangle OPQ is similar to triangle NPR.
(by AAA similarity)