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how to solve for x!!!!!!!!!!!!!!!!!!!!!!!

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8^x-(9)4^x+(26)2^x-24=0

asked Jun 8, 2013 in ALGEBRA 2 by anonymous Apprentice

1 Answer

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The given expression is 8^x-(9)4^x+(26)2^x-24=0

This can be simplified as 2^3x -(9)2^2x +(26)2^x -24 = 0

Assume t=2^x we get,

                                       t^3 -9t^2 +26t -24 = 0   -> (1)

If substitute t value with 2 we will get the answer 0. Hence we conclute that 2 is one of the root.

i.e., 8-36+52-24=0.

Hence synthetic division of the equation (1) by (t-2) we get,

                                      2 | 1  -9   26  -24

                                         |  0   2   -14  -24     

                                             1 -7    12    0

Therefore the equation obtained is t^2 -7t +12 = 0

                                                        =>   t^2 -3t -4t +12 = 0  

                                                        =>   t(t-3) -4(t-4) = 0

                                                        => t=3 and t=4.

We assumed that t=2^x hence we have to take the t value 4 and substitute,

                                                        i.e., 2^x = 4

                                                                2^x = 2^2

                                                          => x=2    [ If bases are equal then we can equalise the exponents ]

Therefore the value of x is 2.

                                                        

answered Jun 8, 2013 by joly Scholar
What about remaining roots for t

The equation is 8x - 9 · 4x + 26 · 2x - 24 = 0 ⇒ (2x)3- 9 · (2x)2 + 26 · (2x) - 24 = 0.

Let 2x = t, then equation t3- 9t2 + 26t - 24 = 0 and solve for t.

The t values are 2, 3 and 4.

If t = 2 then 2x = 21 ⇒ x = 1.

If t = 4 then 2x = 22 ⇒ x = 2.

If t = 3 then 2x = 3 ⇒ x = log(base 2) 3                  image

The solutions of the equation 8x - 9 · 4x + 26 · 2x - 24 = 0 are 1, 2 and log(base 2) 3.

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