If \log_4(x + 2) - \log_{16}(x + 2) = 1, what is the value of x? — JAMB UTME

If $\log_4(x + 2) - \log_{16}(x + 2) = 1$ , what is the value of $x$ ?

Cevap: 14

The value of

x

is 14.

Applying the change of base formula converts

\log_{16}(x + 2)

into

\frac{1}{2}\log_4(x + 2)

. The given equation reduces to

\frac{1}{2}\log_4(x + 2) = 1

, which means

\log_4(x + 2) = 2

. Writing this in exponential form yields

x + 2 = 4^2 = 16

, giving

x = 14

Change the base of

\log_{16}(x + 2)

to base 4

\log_{16}(x + 2) = \frac{\log_4(x + 2)}{\log_4 16} = \frac{1}{2}\log_4(x + 2)

Using the change of base identity

\log_a b = \frac{\log_c b}{\log_c a}

allows all logarithmic terms to share base 4.

Substitute and simplify the expression on the left-hand side

\log_4(x + 2) - \frac{1}{2}\log_4(x + 2) = \frac{1}{2}\log_4(x + 2) = 1

Subtracting half of the term from the whole term leaves half of the term.

Isolate the logarithmic expression

\log_4(x + 2) = 2

Multiplying both sides of the equation by 2.

Convert from logarithmic to index form

x + 2 = 4^2 = 16

By definition,

\log_b a = c \iff b^c = a

Solve for the variable

x

x = 14

Subtracting 2 from both sides gives the final value of

x

Logarithms and Change of Base

Soru