Valaki megoldaná nekem?

1.

log(x) = 1-log(2)

log(x) = log(10)-log(2)

log(x) = log(10/2)

log(x) = log(5)

x = 5

2.

logx + 5 = 14/logx

log^2(x) = 11 // gyököt vonok

log(x) = sqrt(11) // logaritmus alap azonosságát alkalmazva lesz x1 és x2

x1 = 10^sqrt(11)

x2 = 10^-sqrt(11)

-2*logx^-3 + 2*logx^-2 + 3*logx^-4 -10 = 0

-2/logx^3 + 2/logx^2 + 3/logx^4 -10 = 0

-2(logx)/logx^4 + 2(logx^2)/logx^4 + 3/logx^4 = 10

-2(logx)/logx^4 + 2(logx^2)/2(logx^2) + 3/logx^4 = 10

-2(logx)/logx^4 + 1 + 3/logx^4 = 10

-2(logx)/logx^4 + 3/logx^4 = 9

-2(logx)/2(logx^2) + 3/logx^4 = 9

-(logx)/(logx^2) + 3/logx^4 = 9

-1/logx + 3/logx^4 = 9

egyszerűbb alak nem megy

1.

2^(x+1) - 16 = 2^x

2^(x+1) - 2^x = 2^4

2^x = 2^4

x = 4

2.

(5/6)^(3x+3) = (36/25)^(1-x)

(5/6)^(3x+3) = (25/36)^-(1-x)

(5/6)^3*(x+1) = (25/36)^-(1-x)

(5/6)^3*(x+1) = (25/36)^(x-1)

egyszerűbb alak nem megy

3.

3^x - 3 = 54/3^x

3^x - 3 = 54*(1/3^x)

3^x - 3 = 54*3^-x

3^x - 3^1 = 54*3^-x

3^x - 3^1 = (3^3 + 3^3)*3^-x

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

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

egyszerűbb alak nem megy