2^(x^2 - 6x + 7) = 3/4 Clear the denominator. 4 . 2^(x^2 - 6x + 7) = 3 Express 4 as 2^(something). Specifically, 2^2 2^2 . 2^(x^2 - 6x + 7) = 3 Combine the powers by adding them. 2^(x^2 - 6x + 9) = 3 Take log_2 of both sides. x^2 - 6x + 9 = log_2(3) Square root both sides, taking the +/- on the rhs x - 3 = +/- log_2(3) Add 3 x = 3 +/- log_2(3) log_2(3) is about 1.6ish, so the answers are approximately 1.4ish and 4.4ish. You can also use the change of base theorem to write log_2(3) as ln(3)/ln(2) or log(3)/log(2).
I used Log base 2 - and penultimately ended up with x = [6 +- root(4log2(3)) ] / 2 which simplifies to x = 3 +- root(log2(3)). Kicking my self that I didn't see the perfect square.
@@IoT_ Well I had used the standard quadratic formula to solve the problem- but as I said, I there would have been no need for that if I had recognised the initial equation as a perfect square. You've got me intrigued! Please, explain what you mean by 'contracted formula for the even term b'. Your equation looks like the standard quadratic equation but with a couple of typos. I don't understand your final division by just 'a' and not '2a' and why you have just a*c and not 4*a*c inside the square root.
@@johnwilson3918 if b is an even term in ax^2+bx+c=0 , then you can use this formula. We're taught like that in the former USSR republic. I was curious not to hear about it on the Western channels. If b is even then ax^2+bx+c=0 can be represented by ax^2+2k*x+c=0 Then the discriminant is D=b^2-4ac=4k^2-4ac. X1,2= (-b±sqrt(D))/2a= (-2k±sqrt(4k^2-4ac))/2a=(-k±sqrt(k^2-ac))/a Replacing back k=b/2 , we get the formula that I wrote above. It's just kinda easier to use it when b is even.
@@IoT_ Thank you! I feel little embarrassed why I didn't just multiply the top and bottom of your 'contracted equation' by 2 to get back to the standard quadratic. Believe it or not - I was puzzled by your initial expression 'by the even term b'. Had you said something like 'IF b is even' or 'WHEN b is a multiple of 2' (like you did in your last reply)- I would not have been so dismissive. 14 year old me - would be worried about memorising 2 variations ('standard' and 'contracted'). I think it was when I was 17 years old that I heard about 'completing the square' which led me to finding out how the standard quadratic was formulated. There are other methods for solving quadratics - those that are interested - try looking for the MIT review 'A new way to make quadratic equations easy'. Anyhow, let's not forget the fact that using the quadratic formula in any of its variations would not needed if we had recognised that x^2 - 6x + 9 is (x - 3)^2 - ('Square the 1st, Square the Last. Twice the Product'). The 14 year old me could have easily factorised this, Another embarrassing moment for me! I guess I'm getting old.... Best wishes -
@@johnwilson3918 if you meant the method of Po-Shen Loh, then I heard about it. There's even his video here on yt explaining this approach. In my opinion it's just another way to derive a simple quadratic formula. Usually I start with Vieta's formulas , then if it doesn't work I'd proceed with square competition. After that I'd use the formula with the old good discriminant. Best wishes to my fella, an electrical engineer ( I assumed it because of the OR gates IC on your profile picture)
To summarize the change of base thing: Log base a of b = log_a(b) (This is how I'm typing it out) Then, you can rewrite it in terms of another base, c. log_a(b) = log_c(b)/log_c(a) then you can rewrite say, log_3(4) as log_2(4)/log_2(3) = 2/(log_3) If anyone wants to practice a few like this, I've got some sample problems (Don't worry about actually calculating out a numerical approximation; leave your answers in terms of log) Put the following in terms of log_3: log_2(9) log_5(3) log_2.178(89.051) log_9(18) ln(4)
I changed 3/4 to 3.2^-2 and then to 2^(log 3 base 2).2^-2 and then 2^((log 3 base 2) - 2) and equated x^2-6x+7 = (log 3 base 2) - 2 and solved for x using the quadratic formula.
Ummm... it does not work like that... you would also need to divide the square root by ln2 so it would be 3+(sqrt(ln3*ln2)/ln2). Some people like it in a single fraction like in the video and some would do the division. Both ways are correct, simplified answers
@@yogesh193001 how? if you have a sum i the numerator you have to divide all parts of that sum by the denominator. That's why you can simplify the 3ln2 but you have to keep the division of the square root
oh I see the mistake its sqrt(ln3*ln2) not sqrt (ln3/ln2). Just looked at your comment instead of what was in the video :) But still the rest holds up, so I edited the comment
Man, I'm addicted to these videos😂😂😂😂
That's great! Let's spread it 😄😍😁
I am addicted to his σοκολάτα, maths flavors❤❤❤❤❤
Very cool, this!
Perfect.
Thanks 🙏
You’re welcome 😊
Muchas gracias por compartir tan buena explicación de una singular ecuacion exponencial. 😊❤😊.
Np. Thank you for watching!
2^X^2-6x+7=3/4 X=3+-Sqrt[Log[2,3]
Yes. Shift everything from left to right.
x² - 6x + 7 = ln(3/4)/ln(2) := c
x = 3 +/- sqrt (3² - (7 - c))
2^(x^2 - 6x + 7) = 3/4
Clear the denominator.
4 . 2^(x^2 - 6x + 7) = 3
Express 4 as 2^(something). Specifically, 2^2
2^2 . 2^(x^2 - 6x + 7) = 3
Combine the powers by adding them.
2^(x^2 - 6x + 9) = 3
Take log_2 of both sides.
x^2 - 6x + 9 = log_2(3)
Square root both sides, taking the +/- on the rhs
x - 3 = +/- log_2(3)
Add 3
x = 3 +/- log_2(3)
log_2(3) is about 1.6ish, so the answers are approximately 1.4ish and 4.4ish. You can also use the change of base theorem to write log_2(3) as ln(3)/ln(2) or log(3)/log(2).
I used Log base 2 - and penultimately ended up with x = [6 +- root(4log2(3)) ] / 2 which simplifies to x = 3 +- root(log2(3)). Kicking my self that I didn't see the perfect square.
You wouldn't have had to simplify anything if you had used the contracted formula for the even term for b which is:
X1,2=(-b/2 ±sqrt((b/2)^2-a*c))/a
@@IoT_ Well I had used the standard quadratic formula to solve the problem- but as I said, I there would have been no need for that if I had recognised the initial equation as a perfect square. You've got me intrigued! Please, explain what you mean by 'contracted formula for the even term b'. Your equation looks like the standard quadratic equation but with a couple of typos. I don't understand your final division by just 'a' and not '2a' and why you have just a*c and not 4*a*c inside the square root.
@@johnwilson3918 if b is an even term in ax^2+bx+c=0 , then you can use this formula. We're taught like that in the former USSR republic. I was curious not to hear about it on the Western channels.
If b is even then ax^2+bx+c=0 can be represented by
ax^2+2k*x+c=0
Then the discriminant is
D=b^2-4ac=4k^2-4ac.
X1,2= (-b±sqrt(D))/2a= (-2k±sqrt(4k^2-4ac))/2a=(-k±sqrt(k^2-ac))/a
Replacing back k=b/2 , we get the formula that I wrote above.
It's just kinda easier to use it when b is even.
@@IoT_ Thank you! I feel little embarrassed why I didn't just multiply the top and bottom of your 'contracted equation' by 2 to get back to the standard quadratic. Believe it or not - I was puzzled by your initial expression 'by the even term b'. Had you said something like 'IF b is even' or 'WHEN b is a multiple of 2' (like you did in your last reply)- I would not have been so dismissive.
14 year old me - would be worried about memorising 2 variations ('standard' and 'contracted'). I think it was when I was 17 years old that I heard about 'completing the square' which led me to finding out how the standard quadratic was formulated.
There are other methods for solving quadratics - those that are interested - try looking for the MIT review 'A new way to make quadratic equations easy'.
Anyhow, let's not forget the fact that using the quadratic formula in any of its variations would not needed if we had recognised that x^2 - 6x + 9 is (x - 3)^2 - ('Square the 1st, Square the Last. Twice the Product'). The 14 year old me could have easily factorised this, Another embarrassing moment for me! I guess I'm getting old....
Best wishes -
@@johnwilson3918 if you meant the method of Po-Shen Loh, then I heard about it. There's even his video here on yt explaining this approach. In my opinion it's just another way to derive a simple quadratic formula.
Usually I start with Vieta's formulas , then if it doesn't work I'd proceed with square competition. After that I'd use the formula with the old good discriminant.
Best wishes to my fella, an electrical engineer ( I assumed it because of the OR gates IC on your profile picture)
2^(x²-6x+7)=¾
Multiply by 4: 4×2^(x²-6x+7)=3
2²2^(x²-6x+7)=3
2^(x²-6x+9)=3
2^(x-3)²=3
Take log: (x-3)²log(2)=log(3)
x=3±sqrt[{log(3)}/log(2)]
2^-(e/6,55)=0,75
x^2-6x+7=-e/6,55 x^2-6x=-(e/6,55) -7
1,74^2 -6•1,74=-7,415
2^(1,74^2 - 6•1,74+7)=3/4
x=1,74
X^2-6x+7=log(3/4)...x^2-6x+7+2=log3....
To summarize the change of base thing:
Log base a of b = log_a(b) (This is how I'm typing it out)
Then, you can rewrite it in terms of another base, c.
log_a(b) = log_c(b)/log_c(a)
then you can rewrite say, log_3(4) as log_2(4)/log_2(3) = 2/(log_3)
If anyone wants to practice a few like this, I've got some sample problems (Don't worry about actually calculating out a numerical approximation; leave your answers in terms of log)
Put the following in terms of log_3:
log_2(9)
log_5(3)
log_2.178(89.051)
log_9(18)
ln(4)
I changed 3/4 to 3.2^-2 and then to 2^(log 3 base 2).2^-2 and then 2^((log 3 base 2) - 2) and equated x^2-6x+7 = (log 3 base 2) - 2 and solved for x using the quadratic formula.
2^a=3/4
a•ln2=ln3-ln4
a=log₂3-2,
{2^a=2^(log₂3)/2²=3/4}
x²-6x+7=a,
x²-6x+(9-log₂3)=0
{ax²±2kx+c=0 ⇔ x=[∓k±√(k²-ac)]/a}
x=3±√[9-(9-log₂3)]
x=3±√log₂3
@6:12 hold my beer 😂
Divide by ln 2
3+-sqrt(ln3/ln2)
Ummm... it does not work like that... you would also need to divide the square root by ln2 so it would be 3+(sqrt(ln3*ln2)/ln2). Some people like it in a single fraction like in the video and some would do the division. Both ways are correct, simplified answers
@@arkadeusz91you made a mistake
@@yogesh193001 how? if you have a sum i the numerator you have to divide all parts of that sum by the denominator. That's why you can simplify the 3ln2 but you have to keep the division of the square root
oh I see the mistake its sqrt(ln3*ln2) not sqrt (ln3/ln2). Just looked at your comment instead of what was in the video :) But still the rest holds up, so I edited the comment
3ln2/ln2 =3
Sqrt(ln2.ln3)/ln2
=Sqrt(ln2.ln3/ln2.ln2)
=Sqrt(ln3/ln2)
x² - 6x + 9 = log₂3
(x - 3)² = log₂3
*x = 3 ± √log₂3*
Yes, or (6 steps) ...
2^(x² - 6x + 7) = 1/4
2^(x² - 6x + 7) = 2^(log₂(1/4))
2^(x² - 6x + 7) = 2-²
same base (= 2) => x² - 6x + 7 = -2
x² - 6x + 9 = 0
note: x² - 6x + 9 = (x - 3)² x = ±3
/// final results:
■ root #1: x = 3
■ root #2: x = -3
🙂
I used the second method.
Solved in my head in less than 2 minutes: x = 3 +- sqrt(ld(3)) with ld(x) being the base-2 logarithm of x. Not much more than a quadratic equation.
2 ^ (( x - 3 )^2 ) = 3
= 2 ^ (log (3)/log(2))
Hereby
x - 3= √ (log (3)/log(2)),
√ (log (3)/log(2))
x = 3 + √ (log (3)/log(2)),
3 - √ (log (3)/log(2)),
x=(6+/-V(36-4*(7-ln(3/4)/ln2)))/2 , x =~ 4.25895 , 1.74105 , test , x=~ 4.25895 , --> 3/4 ,
x=~ 1.74105 , --> 3/4 , OK ,
Did it worth the electricity wasted?
Left the stupid wsys to solve it
Solving A Cool Exponential Equation: 2^(x² - 6x + 7) = 3/4; x =?
Let: 2ª = 3, a = log₂3 = 1.585; 2^(x² - 6x + 7) = 3/4 = (2ª)/(2²) = 2ª⁻²
x² - 6x + 7 = a - 2, x² - 6x + 9 = a = log₂3, (x - 3)² = (√log₂3)², x - 3 = ± √log₂3
x = 3 ± √log₂3 = 3 ± 1.259; x = 3 + 1.259 = 4.259 or x = 3 - 1.259 = 1.741
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
x = 3 ± √log₂3: x² - 6x + 9 = log₂3; x² - 6x + 7 = log₂3 - 2
2^(x² - 6x + 7) = 2^(log₂3 - 2) = (2^log₂3)(2⁻²) = 3/4; Confirmed
Final answer:
x = 3 + √log₂3 = 4.259 or x = 3 - √log₂3 = 1.741
Hi
Hi