Question

rozdział: funkcje wykladniczne i logartymy

rowiąż równanie

a) 2^x-3 + 2^x = 18

b) 9^x - 25 * 3^x - 54 = 0

c) 81 * (1/3)^x = 9^x/3 (jedna trzecie do x, 9do x przez 3)

d) (3/2)^x * 1,5 = ( 8/27 )^x

e)4^x * 8^x+1 = 1/2

f) 27^x+3 = pierwiastek z 3 * 81^x

g) (4/9)^x * (3/2)^x-1 = 27/8

h) 2,5^x+2 = 0,4 * 0,16^x

proooosze bardzo o rozwiązanie (i jeszcze ktos mogl by jeszcze pare slow wytlumaczenia, z gory dzieki:****)

Answer 1

a)

2^(x-3) + 2^x = 18

2^x  * 2^(-3) + 2 ^x = 18

2^x *( 1/8 + 1) = 18

2^x = 18 : (9/8)

2^x = 18*(8/9) = 16

x = 4

=====

b)

9^x -25 * 3^x - 54 = 0

(3^2)^x - 25* 3^x  - 54 = 0

(3^x)^2 -25* 3^x - 54 = 0

Niech  y = 3^x

y^2 - 25 y - 54 = 0

delta = (-25)^2 - 4*1*(-54) = 625 + 216 = 841

p (delty ) = 29

y = [ 25 - 29]/2 = -4/2 = -2 < 0  - odpada

y = [25 + 29]/2 = 54/2 = 27

czyli

3^x = 27

x = 3

=====

c)

81*(1/3)^x =( 9^x) / 3     / *3

243 * 1/(3^x) =( 3^2)^x

243/ (3^x) = 3^(2x)

243 = 3^(2x) * (3^x)

3^(3x) - 243

3^(3x) = 3^5

3x = 5

x = 5/3

=======

d)

(3/2)^x *1.5 = (8/27)^x

(2/3)^(-x)  *1,5 = (2/3)^(3x)    /  : (2/3)^(-x)

1,5 = (2/3)^[3x -(-x)]

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

(2/3)^(-1)  = (2/3)^(4x)

4x = - 1

x = - 1/4

===========

e)

4^x * 8^(x+1) = 1/2

(2^2)^x * 8* 8^x = 1/2  / : 8

2^(2x)* (2^3)^x  = 1/16

2^(2x)* 2^(3x) = 1/16

2^(5x) = 2^(-4)

5x = - 4

x = -4/5

========

f)

27^(x+3) = p(3)* 81^x

27^x * 27^3 = p(3)* (3^4)^x

(3^3)^x  * (3^3)^3 = 3^(1/2) * 3^(4x)

3^(3x) * 3^9 = 3^(4x + 0,5)  / : 3^(3x)

3^9 = 3^(x +0,5)

9 = x + 0,5

x = 8,5

========

g)

(4/9)^x *(3/2)^(x-1) = 27/8

[(2/3)^2]^x * (3/2)^x * (3/2)^(-1) =  (3/2)^3

(2/3)^(2x) *(2/3) *(3/2)^x = (3/2)^3   / : (3/2)^x

(2/3)^(2x+1) = (3/2)^(3 -x)

(2/3)^(2x +1) = (2/3)^(x -3)

2x + 1 = x -3

x = -3 -1 = -4

============

h)

2,5^(x+2) = 0,4 *0,16^x

(5/2)^(x+2) = (2/5)*(16/100)^x

(5/2)^(x+2) = (2/5)*(4/25)^x

(5/2)^(x+2) = (2/5)*[(2/5)^2]^x

(5/2)^(x+2) = (2/5)*(2/5)^(2x)

(2/5)^(-x -2) = (2/5)^(2x + 1)

-x -2 = 2x + 1

-3 = 3x

x = - 1

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