To solve for x in the equation 1 + log2(3x + 1) = log2(x^2 - 5), we can start by isolating the logarithmic term on one side of the equation.
Subtracting 1 from both sides:
log2(3x + 1) = log2(x^2 - 5) - 1
Now we can use the property of logarithms that states loga(b) = c is equivalent to a^c = b to rewrite the equation without the logarithms:
3x + 1 = 2^(log2(x^2 - 5) - 13x + 1 = 2^(log2(x^2 - 5) - log2(2)3x + 1 = 2^(log2((x^2 - 5)/2)3x + 1 = (x^2 - 5)/2
Now we can solve for x by multiplying both sides by 2 to get rid of the fraction:
6x + 2 = x^2 - 0 = x^2 - 6x - 0 = (x - 7)(x + 1)
Therefore, x = 7 or x = -1.
To solve for x in the equation 1 + log2(3x + 1) = log2(x^2 - 5), we can start by isolating the logarithmic term on one side of the equation.
Subtracting 1 from both sides:
log2(3x + 1) = log2(x^2 - 5) - 1
Now we can use the property of logarithms that states loga(b) = c is equivalent to a^c = b to rewrite the equation without the logarithms:
3x + 1 = 2^(log2(x^2 - 5) - 1
3x + 1 = 2^(log2(x^2 - 5) - log2(2)
3x + 1 = 2^(log2((x^2 - 5)/2)
3x + 1 = (x^2 - 5)/2
Now we can solve for x by multiplying both sides by 2 to get rid of the fraction:
6x + 2 = x^2 -
0 = x^2 - 6x -
0 = (x - 7)(x + 1)
Therefore, x = 7 or x = -1.