1+Log2(3x+1)=Log2(x^2-5)

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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 - 5
0 = x^2 - 6x - 7
0 = (x - 7)(x + 1)

Therefore, x = 7 or x = -1.