If x is a real number such that \log_5(x) + \log_5(x - 20) = 3, what is the value of x? — ACT

If $x$ is a real number such that $\log_5(x) + \log_5(x - 20) = 3$ , what is the value of $x$ ?

Cevap: 25

The value of

x

is 25.

Applying the logarithmic product rule simplifies the equation to

\log_5(x^2 - 20x) = 3

. Writing this in exponential form yields

x^2 - 20x = 125

. Rearranging into standard form gives

x^2 - 20x - 125 = 0

, which factors into

(x - 25)(x + 5) = 0

. This gives potential solutions of

25

and

-5

. Because the logarithmic arguments must be strictly positive,

x = -5

is extraneous. Therefore, the only correct value is 25.

Use the product property of logarithms to combine the terms on the left side.

\log_5(x(x - 20)) = 3

The sum of logarithms with the same base is equal to the logarithm of the product of their arguments:

\log_b(M) + \log_b(N) = \log_b(MN)

Rewrite the logarithmic equation in exponential form.

x(x - 20) = 5^3 \implies x^2 - 20x = 125

The logarithmic equation

\log_b(y) = c

is equivalent to the exponential equation

b^c = y

Rearrange the quadratic equation into standard form and solve by factoring.

x^2 - 20x - 125 = 0 \implies (x - 25)(x + 5) = 0 \implies x = 25 \text{ or } x = -5

Subtracting 125 from both sides sets the quadratic equation to 0, which can then be factored into binomials whose product is 0.

Verify the potential solutions in the original equation to identify any extraneous roots.

For

x = -5

, the arguments of the original logarithms are negative, which is undefined. For

x = 25

, the arguments are positive. Thus, the only valid solution is

x = 25

Logarithmic functions are only defined for positive real numbers. Therefore, we must have

x > 0

and

x - 20 > 0

, which requires

x > 20

Solving logarithmic equations by combining logarithmic terms and checking for extraneous solutions.

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