What are the solutions of the equation logx (2)+log2x (2)=log4x (2)?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll recall the identity: log a (b) = 1/log b (a)

Therefore, using this property, we'll create matching bases in the given equation:

1/ log 2 (x) + 1/log 2 (2x) = 1/log 2 (4x)

We'll recall the product property of the logarithms, such as:

log 2 (2x) = log 2 (2) + log 2 (x)

log 2 (2x) = 1 + log 2 (x)

log 2 (4x) = log 2 (4) + log 2 (x)

log 2 (4x) = 2 + log 2 (x)

We'll substitute log 2 (x) by t:

1/t + 1/(1 + t) = 1/(2+t)

The equation will become:

(1+t)(2+t) + t(2+t) - t(1+t) = 0

We'll remove the brackets:

2 + 3t + t^2 + 2t + t^2 -t - t^2 = 0

We'll eliminate like terms:

t^2 + 3t + 2 + 2t - t = 0

We'll combine like terms:

t^2 + 4t + 2  = 0

We'll apply quadratic formula:

t1 = [-4 + sqrt(16 - 8)]/2

t1 = (-4 + 2sqrt2)/2

t1 = sqrt2 - 2

t2 = -sqrt2 - 2

log 2 (x) = sqrt2 - 2 => x = 2^(sqrt2 - 2)

log 2 (x) = -(sqrt2 + 2) => x = 1/2^(sqrt2 + 2)

Since the x values are positive and they are different from 1, we'll accept them as solutions of the given equation: {2^(sqrt2 - 2) ; 1/2^(sqrt2 + 2)}.

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