Question

Va rog ajutatima oricare din ele macar va rg frumos!!!!

Answer 1

36^x+7·2^x·3^x-78=0
(6)^2x+7-6^x-78=0
notam 6^x=t
t²+7t-78=0
Δ=49+312=361
t1=(-7+19)/2=12/2=6
t2=(-7-19)/2=-26/2=-13
6^x=6
x=1

10·2^x-4^x=16
10·2^x-2^2x-16=0
notam 2^x=t
10t-t²-16=0          (-1)
t²-10t+16=0
Δ=100-64=36
t1=(10-6)/2=4/2=2
t2=(10+6)/2=16/2=8
2^x=2⇒x1=1
2^x=8⇒2^x=2³⇒x2=3
x={1,3}

9^x-10·3^x+9=0
3^2x-10·3^x+9=0
notam 3^x=t
t²-10t+9=0
Δ=100-36=64
t1=(10-8)/2=2/2=1
t2=(10+8)/2=18/2=9
3^x=1⇒3^x=3°⇒x=0
3^x=9⇒3^x=3²⇒x=2
x∈{0,2}

4^x+2^(x+1)=80
2^2x+2^x·2-80=0
notam 2^x=t
t²+2t-80=0
Δ=4+320=324
t1=(-2+18)/2=16/2=8
t2=(-1-18)/2=-20/2=-10
2^x=8⇒2^x=2³⇒x=3

2^2x-2^(x-1/2)=2^(x+1/2)-2^(2x-1)
2^2x-2^x·2^(-1/2)=2^x·2^(1/2)+2^2x·2⁻¹
2^2x-2^x·1/√2=2^x·√2+2^2x·1/2
2^2x-2^x/√2=2^x·√2+2^2x/2
notam 2^x=t
t²-t/√2=√2t+t²/2
t²-t²/2-t/√2-√2t=0
(2t²-t²)/2-√2t/2-√2t=0
t²/2-(√2t+2√2t)/2=0
t²/2-3√2t/2=0             ·2
t²-3√2t=0
t(t-3√2)=0
t1=0
t-3√2=0
t2=3√2
x∈Ф

(0,4)^(x-4)=(6,25)^(4x-7)
0,4=4/10=simplif.cu 2=2/5
6,25=625/100=simplif.cu 25=25/4=(5/2)²=(2/5)⁻²
(2/5)^(x-4)=(2/5)^[-2(4x-7)]
x-4=-2(4x-7)
x-4=-8x+14
x+8x=14+4
9x=18
x=18/9
x=2

(3/11)^(3x-1)=(11/3)^(7x-10)
(3/11)^(3x-1)=(3/11)^-(7x-10)
3x-1=-(7x-10)
3x-1=-7x+10
3x+7x=10+1
10x=11
x=11/10

(4/9)^(x+1)·(27/8)^x=2/3
(2/3)^2(x+1)·(3/2)^3x=2/3
(2/3)^2(x+1)·(2/3)^-3x=(2/3)¹
2(x+1)·(-3x)=1
-6x(2x+1)=1
-12x²-6x-1=0     (-1)
12x²+6x+1=0
Δ=36-48=-12<0
x∈Ф

3^(x+1)+3^x=108
3^x·3+3^x=108
notam 3^x=t
3t+t=108
4t=108
t=108/4
t=27
3^x=27
3^x=3³
x=3

restul nu am inteles ce scrie