A Non-Standard Exponential Equation

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LHS is decreasing, RHS is increasing so there can be only 1 solution, x = 5.

gdtargetvn
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After we saw there is only one solution, we can find it this way:
7^(6-x) = x+2
7^6 = 7^x * (x+2)
u = x+2 ⇒ x = u-2
7^6 = 7^(u-2) * u
7^8 = u * 7^u
7 * 7^7 = u * 7^u
u = 7 ⇒ x = 5

_MATAN_
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7^(6 - x) = x + 2

By observation, x = 5 is a solution since 5 + 2 = 7 = 7^1 = 7^(6-5)

Define f(x) = 7^(6 - x) - x - 2.

f ' (x) = 7^(6 - x) ln 7 (-1) - 1 - 0
f ' (x) = - 7^(6 - x) ln 7 - 1

Note that ln 7 > 0 and 7^(6 - x) > 0 for all x, we see that f ' (x) < 0 for all x.
Thus, f is a decreasing function.
Thus, f(x) = 0 has only one solution, namely x = 5

Packerfan
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y=x+2, then
7*7^7 = y * 7^y

It's easy to show that f(t)= t*7^t is increasing for positive values and negative for t<0, meaning there is a single solution y=7, this means x=5

dmytrovynokurov
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Solution:
7^(6-x) = x+2 |*7^x ⟹
(x+2)*7^x = 7^6 = (5+2)*7^5 |the same operations are done with x on the right side of the equation as with 5 0n the left side of the equation, therefore ⟹
x = 5

gelbkehlchen
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I would suggest performing a substitution, either a = x + 2 or a = 6 - x

then you can put the entire equation in terms of a, which is simpler

for example, if a = 6 - x, then you get:

7^a = 8 - a

which should be a little easier to solve than the original version of the problem

armacham
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Commenting, liked, and subscribe. Great video. I loved the guess and check logic explanation. Thanks!

ChristopherEvenstar
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This is the first one I've ever done in my head. 7^6=(x+2)*7^x, 5 is implied because x+2 can't be anything but 7 in order to satisfy 7-to-the-something on the left, and then, yay, substituting 5 works in 7^1=5+2.

cjburian
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SyberMath's standard solving way! 😏😏😏😏😏😏

alextang
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a million is 10^6, 7 being less than 10, 7^6 is less than a million

malmiteria
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On my method about 7^(6-x)=x+2 we product function
f(x)=7^(6-x) b>1 increased
b/7^x=7^(6-x)= b(1/7)^x
We're derivative function
f'(x)=7^(6-x)×In7>0
Or 7^(6-x)=x+2
log7^(6-x)/log(x+2)=log10
x+2=7
also receive solutions {5, 7} ❤

alinayfeh
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7^6 = 117.649.
Pretty long number, but no... Not even close to 1 million. hehehe

Theuomr
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How would you solve this equation: e^x=x^2-1?

javierpalomino
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It is so simple!
LHS - alwayse decresing
RHS - alwayse increases

So, it has only one real solution.
Gess and check - 5 is a solution.

Ssilki_V_Profile
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For the equation 7^(k-x) = x+2 (assume k>0 and gives a solution) the lambert solution is x=W((ln(7)*(7^(k+2)))/ln(7) - 2. For k=6 we have x=W(ln(7)*(7^8))/ln(7) - 2
ln(7)=1.94591 (approx) and 7^8 =5264801 W(1.94591*5264801)= 13.621371 (approx) 13.621371/1.94591 = (approx). - 2 =

allanmarder