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@GillesF31 15 сағат бұрын
Yes, or ... 7^(x + 8) = 8^(x + 7) 7^x·7^8 = 8^x·8^7 7^8 = (8^x/7^x)·8^7 8^x/7^x = 7^8/8^7 (8/7)^x = 7^8/8^7 x = log(7^8/8^7)/log(8/7) ■ x = 7.572679 🙂
@claudecso62 Күн бұрын
answer is x=4 4^4=256 2^(2×4)=2^8=256
@theodorebartley9776 7 күн бұрын
9Y^2 + 8Y -1 = (9Y-1)(Y+1). No need to use the quadratic formula.
@rajeshpandit8181 10 күн бұрын
Wrong explaination
@rajeshpandit8181 10 күн бұрын
log(a+b)#log a+log b
@stratostrex 11 күн бұрын
The correct answer is x=4.
@milosorevic927 25 күн бұрын
Math olympics for first graders? What is with this yt feed giving me 'olympic' questions that anyone could solge in their head within 3 seconds
@Anthony-tp6fm 26 күн бұрын
You ❤❤❤
@sy8146 28 күн бұрын
Thank you for explaining. From sinx=±1/(√2) , x = ±π/4 (= ± 3.14・・・/4 = ± 0.785・・・ ).
@sy8146 28 күн бұрын
Thank you for explaining. My answer is x = -1, 1. If x=-1, left side is (-1)^(-1) [= (-1)^(odd number) ] = -1 (= right side). Thus, -1 can be one of the solutions.
@oahuhawaii2141 Ай бұрын
You have really bad handwriting. Your '8' is indecipherable, and your rightmost character is either '!' or '1'. I'll do both problems, since I'm bored. For 3!, we have: (a² - 10)/√81 = 3! (a² - 10)/9 = 6 a² = 6*9 + 10 = 64 a = ±8 For 31, we have: (a² - 10)/√81 = 31 (a² - 10)/9 = 31 a² = 31*9 + 10 = 289 a = ±17 It's amazing that both interpretations yield integers. a = ±√(3!*√81+10) = ±8 a = ±√(31*√81+10) = ±17
@oahuhawaii2141 Ай бұрын
I watched the video, and heard you say factorial, so the '1' is really intended to be '!'. BTW, your math isn't rigid in maintaining correct steps. You wrote: a² = 64 √a² = √64 a = ±8 The 2nd line loses the negative sign, but one magically appears in the 3rd line. You should have had written this: a² = 64 a = ±√64 a = ±8 The '√' symbol is the principal square root function, which returns nonnegative values when passed nonnegative values. The '±' symbol is prepended because the inverse of the square function yields 2 values, which can be described in terms of the principal square root and its negated value. Alternatively, you can use these steps: a² = 64 √a² = √64 |a| = 8 a = ±8
@MathElysium19 29 күн бұрын
Welcome bro
@Rando2101 Ай бұрын
Here is what I think: Using AM-GM inequality: 9^sin²(x)+9^cos²(x) >=2sqrt(9)=6 Which means the equality happens when sin(x)=cos(x) You can continue from there to get the solution
@Sergey_Moskvichev Ай бұрын
(7/8)ˣ=8⁷/7⁸=8⁷/(7⁷×7)=(8/7)⁷×1/7=(7/8)⁻⁷×1/7. Прологарифмируйте по основанию 7/8 получите x=log(1/7)-7. Такой ответ красивый.
@SidneiMV Ай бұрын
piece of cake. 4^4 = x^x . *x = 4*
@hangthuy458 Ай бұрын
a^2-10/(81)^1/2=3! a^2-10/9=6 a^2=9×6+10=54+10=64 a=+-8
@RealQinnMalloryu4 Ай бұрын
W^{9sinx^2+9sinx^2➖ +9cosx^2+9cosx^2 ➖}^w =w^{18sinx^4+18cosx^4^}^w= w^{36sinxcosx^8}^w w^{6^6sinxcosx^4}^w w^{3^2^3^2sinxcosx^2^2}^w w^{1^1^3^1sinxcosx^1^2}^w =(w^{3sinxcosx^2}^w).
@robdavies2483 Ай бұрын
Doesn't X = PI/4 OR 0.78...
@walterwen2975 Ай бұрын
x^x = x, (x^x)/x = 1, x^(x - 1) = 1 = x^0; x - 1 = 0; x = 1
@DedMatveev Ай бұрын
There must be 3 solutions, one to the left of zero, because the power 4 is even, and two to the right, because the exponential function grows faster than the power function.
@punjabitehzeeb6725 4 ай бұрын
👍🏻