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Solve the equation log(base 2)(x) + log(base 4)(x+1) = 3.​

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User REJH
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8.1k points

1 Answer

3 votes

We can use the logarithmic identity log_a(b) + log_a(c) = log_a(bc) to simplify the left side of the equation:

log_2(x) + log_4(x+1) = log_2(x) + log_2((x+1)^(1/2))

Using the rule log_a(b^c) = c*log_a(b), we can simplify further:

log_2(x) + log_2((x+1)^(1/2)) = log_2(x(x+1)^(1/2))

Now we can rewrite the equation as:

log_2(x(x+1)^(1/2)) = 3

Using the rule log_a(b^c) = c*log_a(b), we can rewrite this as:

x(x+1)^(1/2) = 2^3

Squaring both sides, we get:

x^2 + x - 8 = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 1, and c = -8. Plugging in these values, we get:

x = (-1 ± sqrt(1^2 - 4(1)(-8))) / 2(1)

x = (-1 ± sqrt(33)) / 2

x ≈ -2.54 or x ≈ 3.54

However, we must check our solutions to make sure they are valid. Plugging in x = -2.54 to the original equation results in an invalid logarithm, so this solution is extraneous. Plugging in x = 3.54 yields:

log_2(3.54) + log_4(4.54) = 3

0.847 + 0.847 = 3

So x = 3.54 is the valid solution to the equation.

answered
User Adam Tegen
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9.5k points

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