Find the value of x that satisfies the equation log_2(x²-9)=3log_2 2+1
Starting with the equation:
log₂(x² - 9) = 3log₂(2) + 1
We can simplify the right side by using the property of logarithms that states loga(b^c) = cloga(b).
This allows us to rewrite 3log₂(2) as log₂(2³), which equals log₂(8).
Our equation now becomes:
log₂(x² - 9) = log₂(8) + 1
Using the property that loga(b) = loga(c) if and only if b = c, we can eliminate the logarithms and set the expressions inside equal to each other:
x² - 9 = 8 + 1
Combining like terms:
x² - 9 = 9
Adding 9 to both sides:
x² = 18
Taking the square root of both sides:
x = ±√18
Simplifying the square root of 18:
x = ±√(9 * 2)
x = ±3√2
Therefore, the values of x that satisfy the equation are x = 3√2 and x = -3√2.
To find the value of x that satisfies the equation log₂(x² - 9) = 3log₂ 2 + 1, we can start by simplifying the right side of the equation.
Recall that log₂ 2 = 1 because any number raised to the power of 1 is itself. So, we can rewrite the equation as:
log₂(x² - 9) = 3 + 1
Simplifying further:
log₂(x² - 9) = 4
Now, we can match the equation with the logarithmic definition. In logarithmic form, logₐ b = c can be written as aᶜ = b.
Applying this to our equation, we have:
2^4 = x² - 9
Simplifying further:
16 = x² - 9
Rearranging the equation:
x² = 16 + 9
x² = 25
Taking the square root of both sides:
x = ± √25
So, the value of x that satisfies the equation is x = ±5.
To find the value of x that satisfies the equation:
log₂(x²-9) = 3log₂2 + 1
Step 1: Simplify the right-hand side of the equation.
3log₂2 + 1 = 3(1) + 1 [Using the property log₂a = 1, where a is any positive number]
= 3 + 1
= 4
So, the equation becomes:
log₂(x²-9) = 4
Step 2: Rewrite the equation in exponential form.
2⁴ = x²-9
Step 3: Simplify the left-hand side.
16 = x²-9
Step 4: Rearrange the equation.
x² = 16 + 9
x² = 25
Step 5: Take the square root of both sides to solve for x.
√(x²) = √25
x = ±√25
Therefore, the values of x that satisfy the equation are x = -5 and x = 5.