Please help me solve for x

Please help me solve for x

Postby juriemagic » Fri Oct 23, 2015 7:16 am

-3.2^(x+7)+3.2^(x+9)=36

I have been struggling for a while with this one, please help..
juriemagic
 
Posts: 2
Joined: Fri Oct 23, 2015 7:11 am
Reputation: 0

Re: Please help me solve for x

Postby Guest » Fri Oct 23, 2015 11:57 am

The key here is [tex]2^{x+9} = 2^{x+7 + 2} = 2^{x+7}\cdot2^2[/tex] also delete the both sides by 3

Can you continue by yourself?
Guest
 

Re: Please help me solve for x

Postby Guest » Fri Oct 23, 2015 7:31 pm

-3.2^(x+7)+3.2^(x+9)=36
OR another approach would be........
keep it as simple as possible first divide across by 3

-2^(x+7) + 2^(x+9)=12 ...then re-arrange

2^(x+9) = 12 + 2^(x+7) ...then divide across by 2^(x+9)

1 = [12 / 2^(x+9)] + [2^(x+7) / 2^(x+9)]

1 = [12 / 2^(x+9)] + [2^(x+7-x-9)] ...subtract indices

1 = [12 / 2^(x+9)] + [2^(-2)] .....2 to power -2 is a quarter

1 = [12 / 2^(x+9)] + [1/4] .....subtract the 1/4 both sides

3/4 = [12 / 2^(x+9)] ....multiply by 4 across

3 = [48 / 2^(x+9)] ...... divide by 3 across

1 = [16 / 2^(x+9) ...cross multiply

2^(x+9) = 16 = 2^4

therefore equating indices .... X + 9 = 4 OR X = -5
Guest
 

Re: Please help me solve for x

Postby Guest » Sat Oct 24, 2015 3:43 pm

-3.2^(x+7)+3.2^(x+9)=36

-3.2^(x+7)+3.4.2^(x+7)=9.2^2

-3.2^(x+7)+12.2^(x+7)=9.2^2

9.2^(x+7)=9.2^2

X+7=2

X=-5
Guest
 

Re: Please help me solve for x

Postby juriemagic » Mon Nov 09, 2015 10:57 am

To all of those who have helped, please accept my apologies for not replying promptly. I will have a look at these answers tomorrow. Thank you all very kindly.

juriemagic
 
Posts: 2
Joined: Fri Oct 23, 2015 7:11 am
Reputation: 0

Re: Please help me solve for x

Postby leesajohnson » Mon Aug 29, 2016 6:50 am

=> -3.2^(x+7)+12.2^(x+7) = 9.2^2

=> 9.2^(x+7) = 9.2^2

=> X+7 = 2

=> X = -5
User avatar
leesajohnson
 
Posts: 208
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33


Return to Polynomials, Polynomial Identities



Who is online

Users browsing this forum: No registered users and 1 guest