C1 Indices Question HELP!

Hi,

Please Could anyone help (with method) how to solve for x (answer is 2)

2^2x - 6(2^x+1) + 32 = 0
(2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

Thanks for the help!

Reply 1

RDKGames

Study Forum Helper

Original post by umbrocurry

Hi,

Please Could anyone help (with method) how to solve for x (answer is 2)

2^2x - 6(2^x+1) + 32 = 0
(2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

Thanks for the help!

So....

2^{2x}-6(2^{x+1})+32=0

Reply 2

RDKGames

Study Forum Helper

Original post by umbrocurry

Hi,

Please Could anyone help (with method) how to solve for x (answer is 2)

2^2x - 6(2^x+1) + 32 = 0
(2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

Thanks for the help!

You should know that

2^{x+1}=2^x \cdot 2^1

Your equation becomes:

(2^x)^2-6(2)(2^x)+32=0 \Rightarrow (2^x)^2-12(2^x)+32=0

which you can solve for

2^x

and then

x

on its own.

x=2

is not the only answer.

Reply 3

NotNotBatman

2^{2x}-6(2^{x+1})+32=0

Remember that

a^{b+c} = a^b \times a^c

2^{x+1}

can be written as

2 \times 2^x

Reply 4

Daffy786

Original post by RDKGames

You should know that

2^{x+1}=2^x \cdot 2^1

Your equation becomes:

(2^x)^2-6(2)(2^x)+32=0 \Rightarrow (2^x)^2-12(2^x)+32=0

which you can solve for

2^x

and then

x

on its own.

x=2

is not the only answer.

Yes but how can I solve as I can't get all the base numbers equal- 32 is 2^5 but then I am left with 12??

Reply 5

RDKGames

Study Forum Helper

Original post by Daffy786

Yes but how can I solve as I can't get all the base numbers equal- 32 is 2^5 but then I am left with 12??

What does that even mean?? Just solve it as you would solve any quadratic - factorise it, or complete the square if needed.

Reply 6

umbrocurry

Original post by RDKGames

What does that even mean?? Just solve it as you would solve any quadratic - factorise it, or complete the square if needed.

yes but how exactly?

Reply 7

RDKGames

Study Forum Helper

Original post by umbrocurry

yes but how exactly?

Let

y=2^x

\Rightarrow y^2-12y+32=0

Then when you have y=something just replace y with

2^x

and take logs with base 2 of both sides to get

x

Reply 8

umbrocurry

Original post by RDKGames

Let

y=2^x

\Rightarrow y^2-12y+32=0

Then when you have y=something just replace y with

2^x

and take logs with base 2 of both sides to get

x

Oh Yeah!!! Thanks so much! I forget substitution and I finally got the answer now.

On a side note: why is (2^x)^2 equal to 2^2x and not 4^2x, since surely you would have to square the integer too?

Thanks alot :wink: