Mathematician's Please Help With The Following Equations - Education

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Mathematician's Please Help With The Following Equations by Samwils(op): 4:20pm On Mar 14, 2019

Below

Re: Mathematician's Please Help With The Following Equations by Samwils(op): 4:44pm On Mar 14, 2019

Please I need help

Re: Mathematician's Please Help With The Following Equations by Samwils(op): 4:52pm On Mar 14, 2019

Please who can solve it oh.. I'm confused

Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:01pm On Mar 14, 2019

no 1,x=2
no 2, x=-4

Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:03pm On Mar 14, 2019

no 2 is easier, rearrange it u have
x=log0.0001 to base10
that's -4

Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:10pm On Mar 14, 2019

for no 1, the solution is longer. In summary
Use laws of indices to express in terms of 3^x. Make sure 3^x appears in all the terms
Equate 3^x to p,
solve quadratic equation in terms of P, you get P= -1 or 9, It's 9 because P(representing 3^x) must be positive
If 3^x=9, then x=2 because 3^2 is 9.
I'd have sharply written and snapped but my camera no good.
I suffered a lot during my days of further maths.
This was one of the topics I still enjoyed wink

Re: Mathematician's Please Help With The Following Equations by dejt4u(m): 5:21pm On Mar 14, 2019*

Samwils:
Below

No 1
3^2(x-1) - 8(3^x-2) = 1,
3^2x-2 - 8(3^x-2) = 1,
(3^2x ÷ 3²) - 8(3^x ÷ 3²) = 1,
let y = 3^x,
so we have;
(y² ÷ 9) - 8(y ÷ 9) = 1,
y² - 8y - 9 = 0,
solve quadratically

No 2.
10^x = 0.0001,
10^x = 10^-4,
x = -4

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Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:23pm On Mar 14, 2019

dejt4u:
No 1
3^2(x-1) - 8(3^(x-1)) = 1,
3^2x-2 - 8(3^(x-1)) = 1,
(3^2x ÷ 3²) - 8(3^x ÷ 3¹) = 1,
let y = 3^x,
so we have;
(y² ÷ 9) - 8(y ÷ 3) = 1,
y² - 24y - 9 = 0,
solve quadratically

No 2.
10^x = 0.0001
10^x = 10^-4
x = -4

How do you type powers like that?

2 Likes

Re: Mathematician's Please Help With The Following Equations by dejt4u(m): 5:34pm On Mar 14, 2019

Honeyprof:
How do you type powers like that?

use this [sup]power here[/sup*]

pls remove the * in the parenthesis

1 Like

Re: Mathematician's Please Help With The Following Equations by Martinez39(m): 8:21pm On Mar 14, 2019

1) 3^{2(x - 1)} - 8( 3^{(x - 2)} ) = 1
Using the laws of indices and distributive property of real numbers,
( 3^2x / 3²) - 8( 3^x / 3²) = 1
( 3^2x ) - 8 ( 3^x) = 9
( 3^x )² - 8 ( 3^x) - 9 = 0
Let u = 3 ^x
u² - 8u - 9 = 0, solving this quadratic equation using quadratic formula yields u = 9 & -1 are solutions.
3^x = 9 = 3² implies x = 2 is a solution.
3^x = -1 has no real solution.
Therefore, x = 2

2) 10^x = 0.0001 = 10^-4
Taking the common logarithm of both sides,
log 10^x = log 10^-4
x log 10 = -4 log 10 (using properties of logarithm)
Therefore, x = -4

3 Likes

Re: Mathematician's Please Help With The Following Equations by MKULTRA: 8:26pm On Mar 14, 2019

Petition for maths should be banned angry

Re: Mathematician's Please Help With The Following Equations by dejt4u(m): 9:56pm On Mar 14, 2019

Martinez39:
1) 3^{2(x - 1)} - 8( 3^{(x - 2)} ) = 1
Using the laws of indices and distributive property of real numbers,
( 3^2x / 3²) - 8( 3^x / 3²) = 1
( 3^2x ) - 8 ( 3^x) = 9
( 3^x )² - 8 ( 3^x) - 9 = 0
Let u = 3 ^x
u² - 8u - 9 = 0, solving this quadratic equation using quadratic formula yields u = 9 & -1 are solutions.
3^x = 9 = 3² implies x = 2 is a solution.
3^x = -1 has no real solution.
Therefore, x = 2

2) 10^x = 0.0001 = 10^-4
Taking the common logarithm of both sides,
log 10^x = log 10^-4
x log 10 = -4 log 10 (using properties of logarithm)
Therefore, x = -4

you're correct

Re: Mathematician's Please Help With The Following Equations by Here2day(m): 10:54pm On Mar 14, 2019

Martinez39:
1) 3^{2(x - 1)} - 8( 3^{(x - 2)} ) = 1
Using the laws of indices and distributive property of real numbers,
( 3^2x / 3²) - 8( 3^x / 3²) = 1
( 3^2x ) - 8 ( 3^x) = 9
( 3^x )² - 8 ( 3^x) - 9 = 0
Let u = 3 ^x
u² - 8u - 9 = 0, solving this quadratic equation using quadratic formula yields u = 9 & -1 are solutions.
3^x = 9 = 3² implies x = 2 is a solution.
3^x = -1 has no real solution.
Therefore, x = 2

2) 10^x = 0.0001 = 10^-4
Taking the common logarithm of both sides,
log 10^x = log 10^-4
x log 10 = -4 log 10 (using properties of logarithm)
Therefore, x = -4

Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:02pm On Mar 15, 2019

dejt4u:
use this [sup]power here[/sup*]

pls remove the * in the parenthesis

Thank you

Re: Mathematician's Please Help With The Following Equations by Honeyprof: 5:05pm On Mar 15, 2019

Martinez39:
1) 3^{2(x - 1)} - 8( 3^{(x - 2)} ) = 1
Using the laws of indices and distributive property of real numbers,
( 3^2x / 3²) - 8( 3^x / 3²) = 1
( 3^2x ) - 8 ( 3^x) = 9
( 3^x )² - 8 ( 3^x) - 9 = 0
Let u = 3 ^x
u² - 8u - 9 = 0, solving this quadratic equation using quadratic formula yields u = 9 & -1 are solutions.
3^x = 9 = 3² implies x = 2 is a solution.
3^x = -1 has no real solution.
Therefore, x = 2

2) 10^x = 0.0001 = 10^-4
Taking the common logarithm of both sides,
log 10^x = log 10^-4
x log 10 = -4 log 10 (using properties of logarithm)
Therefore, x = -4

1 Reply

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