Solving Simultaneous Equations and Understanding Exponential Properties

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The discussion revolves around solving simultaneous equations and understanding exponential properties. The first equation, y = x - 1 and y = x^2 - 3, was clarified by suggesting to substitute for y instead of x. The second equation, 72^x = 49, was approached by expressing 49 in terms of base 7, leading to the conclusion that x = 1. Additionally, the conversation highlighted the importance of manipulating bases in exponential equations, with examples provided for clarity. Ultimately, the participants worked through the problems collaboratively, emphasizing the principles of exponentiation and substitution.
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Just doing a bit of math homework and I am stuck on 2 questions

the first one is a similtanoius equation and goes like this

y = x - 1
y = x2 -3

ive been trying for about half an hour and all i end up with is

y = (square root of) y - 2

im not sure if it's right, it does not look it.

the 2nd question i don't know the name of and looks like

72x = 49

i don't even know where to begin on that, i just have to factorise it.

thankyou in advanced
 
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3nder said:
y = x - 1
y = x2 -3

ive been trying for about half an hour and all i end up with is

y = (square root of) y - 2

Instead of substituting for x, substitute for y instead.


3nder said:
the 2nd question i don't know the name of and looks like

72x = 49

Can you write 49 in any way that relates to 7?
 
for the 2nd question i know that x = 1 but i don't know how to write the proof, like in another question how would i do

43-x = 8x

1st question solved thankyou rock.freak667
 
Last edited:
3nder said:
for the 2nd question i know that x = 1 but i don't know how to write the proof, like in another question how would i do

43-x = 8x

It all revolves around how you can write the base numbers. If you have am=an then m=n.

So if you have 36=62x, then you can write this as 62=62x which means that 2=2x and hence x=1.
 
3nder said:
for the 2nd question i know that x = 1 but i don't know how to write the proof, like in another question how would i do

43-x = 8x

1st question solved thankyou rock.freak667
Also remember how a power to a power behaves:
(a^b)^c = a^{bc}

so:
(8)^x=(2^3)^x=2^{3x}

and

(4)^{3-x}=(2^2)^{3-x}=2^{6-2x}

edit: I misread and thought you said this problem's solution was x = 1. My bad!
 
xcvxcvvc said:
Also remember how a power to a power behaves:
(a^b)^c = a^{bc}

so:
(8)^x=(2^3)^x=2^{3x}

and

(4)^{3-x}=(2^2)^{3-x}=2^{6-2x}

edit: I misread and thought you said this problem's solution was x = 1. My bad!

Right so now you have

2^{6-2x}=2^{3x}

so what is x?
 
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