1). change to exponential form:
log2^256=8; 2^8=256
log0.0001=-4; 10^-4=0.0001
2). change to logarithmic form:
5^3=125; log5^125=3
4^-3=1/64; log4^1/64=-3
3). solve the equation:
3^(x-2)=27^(x+1)
3^(x-2)=(3^3)^(x+1)
3^(x-2)=3^(3x+3)
x-2=3x+3
-5=2x
-5/2=x
2^(x^(2)+1)=4^(x+2)
2^(x^(2)+1)=(2^2)^(x+2)
2^(x^(2)+1)=2^(2x+4)
x^(2)+1=2x+4
x^(2)-2x-3
(x+1)(x-3)
x=-1;x=3
all correct.
The solutions to the equation 3^(x-2) = 27^(x+1) are x = -1 and x = 3.
To solve the equation 3^(x-2) = 27^(x+1), we can use the fact that 27 is equal to 3 raised to the power of 3.
So we can rewrite the equation as 3^(x-2) = (3^3)^(x+1).
Applying the exponent property, we have 3^(x-2) = 3^(3*(x+1)).
Since the bases are equal, we can equate the exponents: x - 2 = 3*(x + 1).
Expanding the equation, we get x - 2 = 3x + 3.
Bringing the x terms to one side and the constants to the other side, we have x - 3x = 3 + 2.
Simplifying the equation, we get -2x = 5.
Dividing both sides by -2, we find x = -5/2.
To solve the equation 2^(x^2 + 1) = 4^(x + 2), we can rewrite the equation using the fact that 4 is equal to 2 raised to the power of 2.
So we have 2^(x^2 + 1) = (2^2)^(x + 2).
Simplifying further, we get 2^(x^2 + 1) = 2^(2*(x + 2)).
Since the bases are equal, we can equate the exponents: x^2 + 1 = 2*(x + 2).
Expanding the equation, we have x^2 + 1 = 2x + 4.
Bringing the x terms to one side and the constants to the other side, we get x^2 - 2x - 3 = 0.
The equation is now a quadratic equation, so let's factor it:
(x + 1)(x - 3) = 0.
Using the zero-product property, either (x + 1) = 0 or (x - 3) = 0.
Solving each equation gives us x = -1 or x = 3.