Exponencialní rovnice, příklady
Jak se řeší priklad 1,3? Děkuji,urcuje se jeste u toho něco?
Odpovědi
Diskuze
4^x - 80/(4^x) = -16
80/(4^x) - 4^x = 16
80 - 4^x * (4^x) = 16 * (4^x)
substituce: w = 4^x
80 - w^2 = 16 * w
0 = w^2 + 16 * w - 80; kvadratická rovnice
3) zadání není čitelné
https://www.wolframalpha.com/input/?i=0+%3D+w%5E2+%2B+16+*+w+-+80
w1 = 4^x1
4 = 4^x1
x1=1
====
zkouška:
4 - 80/4 = -16
---
-20 = 4^x2; nemá řešení v R; druhé řešení je imaginární
2)
5^x + 3*2^(x+2) = 5^-2 * 5^(x+3) + 2^(x+2)
5^x + 3*2^(x+2) = 1/5^2 * 5^x * 5^3 + 2^(x+2)
5^x + 3*2^(x+2) = 5 * 5^x + 2^(x+2)
2^(x+2)*(3-1) = (5-1) * 5^x
2^(x+2)*(2) = (4) * 5^x
2^(x+2) / 2 = 5^x
2^(x+2-1) = 5^x
ln 2^(x+1) = ln 5^x
1 = x*ln(5)/ln(2) - x
1 = x*(ln(5)/ln(2) - 1)
1/(ln(5)/ln(2) - 1) = x
x=1/(ln(5)/ln(2) - 1)
===
2^(3x+2)/16^x * 8 = (1/4)^(x-1) * 32
2^(3x+2)/16^x = (1/4)^(x-1) * 4
2^(3x+2)/16^x = 4^(1-x) * 4
2^(3x+2) = 16 * 16^x/4^x
2^(3x+2) = 16 * (4^2)^x/4^x
2^(3x+2) = 16 * 4^(2x)/4^x
2^(3x+2) = 16 * 4^(2x-x)
2^(3x+2) = 2^4 * (2^2)^x
2^(3x+2) = 2^4 * 2^(2x)
2^(3x+2) = 2^(2x + 4)
(3x+2) = (2x + 4)
x = 2
=====
Zkouška:
2^(3x+2)/16^x * 8 = (1/4)^(x-1) * 32
2^(3*2+2)/16^2 * 8 = (1/4)^(2-1) * 32
2^(8)/(2^4)^2 = 1/4 * 4
2^(8) / 2^(2*4) = 1/4 * 4
1 = 1; OK
4)
0,25^(2-x) = 256/ 2^(x+3)
(1/4)^(2-x) = 256/ 2^(x+3)
4^(x-2) = 256/ 2^(x+3)
(2^2)^(x-2) = 256/ 2^(x+3)
2^(2x-4) = 256/ 2^(x+3)
2^(2x-4) * 2^(x+3) = 256
2^(2x-4 + x+3) = 2^8
2^(3x-1) = 2^8
(3x-1) = 8
3x = 9
x=3
===
5)
(1/3)^(x-1) * 9^x = 27 * (1/81) ^ (1-x)
3^(1-x) * 9^x = 27 * (81) ^ (x-1)
3^(1-x) * (3*3)^x = 27 * (9*9) ^ (x-1)
3^(1-x) * 3^2x = 3^3 * (3^4) ^ (x-1)
3^(1-x) * 3^2x = 3^3 * 3^(4x-4)
3^(1-x + 2x) = 3^(4x-4 + 3)
3^(1+x) = 3^(4x-1)
(1+x) = (4x-1)
2 = 3x
x=2/3
=====
6)
15*2^(x+1) + 15 * 2^(-x+2) = 135
15*2^(x+1) + 15 * 2^(-x+2) = 15*9
2^(x+1) + 2^(-x+2) = 9
2^(x) + 2^(-x+1) = 9/2
(2^x)^2 + 2 = 9/2 * 2^x
(2^x)^2 - 9/2 * 2^x + 2 = 0
y=2^x
y^2 - 9/2 * y + 2 = 0
y1=1/2
y2=4
1/2=2^x1
log2(1/2)=log2(2^x1)
x1=-1
=====
4=2^x2
log2(4)=log2(2^x2)
x2=2
====
7)
6^x - 4*3^x = 3*2^x - 12
(2*3)^x - 4*3^x = 3*2^x - 12
2^x * 3^x - 4*3^x = 3*2^x - 3*4
(2^x - 4)*3^x = 3*(2^x - 4)
3^x = 3
x=1
===
8)
4^x - 10*2^(x-1) = 24
(2^2)^x - 10*2^(x-1) = 24
(2^x)^2 - 5*2^x - 24 = 0
subst: y=2^x
y^2 - 5*y - 24 = 0
y1=-3
x2=8
-3=2^x1
x1=nemá řešení
8=2^x2
x2=3
====
9)
5*4^(x+1) - 4^(x+2) = 4^(x-1) + 240
5*4^(x+2) - 4^(x+3) = 4^(x) + 240*4
5*4^2 * 4^x - 4^3 * 4^x = 4^x + 240*4
(5*16 - 16*4 - 1)*4^x = 240*4
4^x = 240*4/(5*16 - 16*4 - 1) = 64
x=3
===