What value of the base b satisfies the equation \log_b (3b + 10) = 2? — ACT

What value of the base $b$ satisfies the equation $\log_b (3b + 10) = 2$ ?

Answer: 5

The correct answer is 5.

Applying the definition of logarithms to the equation

\log_b(3b + 10) = 2

converts it to the quadratic equation

b^2 = 3b + 10

. Subtracting

3b

and

10

from both sides results in

b^2 - 3b - 10 = 0

. Factoring the quadratic yields

(b - 5)(b + 2) = 0

, giving the potential solutions

b = 5

and

b = -2

. Because the base of a logarithm must be positive (

b > 0

), the negative solution is extraneous. This leaves

5

as the only valid base.

Convert the logarithm to exponential form.

b^2 = 3b + 10

By the definition of logarithms,

\log_b(x) = y

is equivalent to

b^y = x

Rearrange into standard quadratic form.

b^2 - 3b - 10 = 0

Subtracting

3b

and

10

from both sides sets the quadratic expression equal to zero.

Factor the quadratic equation.

(b - 5)(b + 2) = 0

Finding two integers that multiply to

-10

and add to

-3

gives

-5

and

2

Solve for the variable and apply base constraints.

b = 5

Solving the factored equation yields

b = 5

b = -2

. Since a logarithmic base must be strictly positive (

b > 0

), we reject

b = -2

as extraneous, leaving

b = 5

Converting logarithmic equations to exponential form and verifying base restrictions.

Estimated Time:1m 30s

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