If x and y are positive real numbers greater than 1 such that \log_y(x) + 6\log_x(y) = 5 a… — ACT

If $x$ and $y$ are positive real numbers greater than $1$ such that $\log_y(x) + 6\log_x(y) = 5$ and $xy = 64$ , what is the sum of all possible values of $x$ ?

$16 + 16\sqrt{2}$ Cevap
B
$4 + 2\sqrt{2}$
C
$4\sqrt{2}$
D
$80$
E
$528$

Cevap

16 + 16\sqrt{2}

Using the change of base property

\log_x(y) = \frac{1}{\log_y(x)}

, the equation can be written in terms of

u = \log_y(x)

u + \frac{6}{u} = 5

. Solving this quadratic equation gives

u = 2

u = 3

. If

\log_y(x) = 2

, then

x = y^2

. Substituting into

xy = 64

yields

y^3 = 64 \implies y = 4

, which gives

x = 16

. If

\log_y(x) = 3

, then

x = y^3

. Substituting into

xy = 64

yields

y^4 = 64 \implies y = 2\sqrt{2}

, which gives

x = 16\sqrt{2}

. The sum of these values is

16 + 16\sqrt{2}

Adım Adım Çözüm

Apply the change of base formula to express the equation in terms of a single logarithmic base.

Since

\log_x(y) = \frac{1}{\log_y(x)}

, the equation

\log_y(x) + 6\log_x(y) = 5

becomes

\log_y(x) + \frac{6}{\log_y(x)} = 5

This allows us to solve for the log expression using substitution.

Substitute

u = \log_y(x)

and solve the resulting quadratic equation for

u

u + \frac{6}{u} = 5 \implies u^2 - 5u + 6 = 0 \implies (u-2)(u-3) = 0

, so

u = 2

u = 3

Solving the quadratic equation gives the possible relationships between

x

and

y

Analyze the first case where

u = 2

and solve for

x

\log_y(x) = 2

, then

x = y^2

. Substituting this into

xy = 64

gives

y^3 = 64

, which yields

y = 4

. Thus,

x = 4^2 = 16

This determines the first possible value of

x

Analyze the second case where

u = 3

and solve for

x

\log_y(x) = 3

, then

x = y^3

. Substituting this into

xy = 64

gives

y^4 = 64

, which yields

y = 64^{1/4} = (2^6)^{1/4} = 2^{3/2} = 2\sqrt{2}

. Thus,

x = (2\sqrt{2})^3 = 16\sqrt{2}

This determines the second possible value of

x

Sum the possible values of

x

16 + 16\sqrt{2}

The question asks for the sum of all possible values of

x

Anahtar Kavram

Solving systems of exponential and logarithmic equations using base-change properties and substitution

Alternatif Yöntem

Instead of using substitution directly, you can write both equations in terms of base 2 or natural logs: let

\log_y(x) = k

, which means

x = y^k

. We then have

k + 6/k = 5

giving

k = 2

k = 3

. This leads directly to

x = y^2

and

x = y^3

, which can then be substituted into the second equation.

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