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Source: — Problem Solving |
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by KeyserSoze525 » Mon Oct 27, 2008 6:49 am
2^x - 2^(x-2) = 3(2^13)

Using the exponent rule x^a * x^b = x^(a+b):

2^x - 2^x*2^-2 = 3(2^13)

Factor out 2^x from the left side:

2^x (1 -2^-2) = 3(2^13)

Simplify:

2^x * (1 - 1/4) = 3(2^13)

2^x * 3/4 = 3(2^13)

Multiply both sides by 4 or 2^2:

2^x * 3 = 2^2 * 3(2^13)

Divide both sides by 3:

2^x = 2^2 * 2^13

Simplify:

2^x = 2^(2+13)

2^x = 2^15

Since the bases are the same you can say that x = 15
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by raunekk » Thu Oct 30, 2008 1:45 am
imo:15
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by vivek.kapoor83 » Thu Oct 30, 2008 3:00 am
clearly 15.
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by lachlanc » Thu Oct 30, 2008 8:25 am
vivek.kapoor83 wrote:clearly 15.
what value did you bring to the discussion with that statement?
If it so clear that the answer is 15, please show your work.
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by PAB2706 » Fri Oct 31, 2008 12:25 am
2^x - 2^(x-2) = 3(2^13)

let us first solve the LHS

= 2^x - 2^x/ 2^2 solving by equating the denominator we get

= (2^x* 2^2 - 2^x) / 2^2

= [2^x * ( 2^2-1)] / 2^2

= [2^x * ( 4-1 ) ] / 2^2

= [ 2^x * ( 3 ) ] / 2^2

Now equating LHS and RHS we get

2^x * ( 3 ) = 3* (2^13 * 2^2)

2^x * ( 3) = 3* 2^15

cancelling 3 from both the sides and then comparing the exponents of 2 we get

x=15

I know this is too elaborate explanation but i am sure this shud clear most doubts.
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