Find the value of x that satisfies the exponential equation \frac{4^{x+1} + 4^x + 4^{x-1}}… — JAMB UTME

Find the value of

x

that satisfies the exponential equation

\frac{4^{x+1} + 4^x + 4^{x-1}}{2^{x+2} + 2^{x+1} + 2^x} = 24

Answer: 5

Factoring

4^{x-1}

from the numerator yields

4^{x-1}(16 + 4 + 1) = 21 \cdot 4^{x-1}

. Factoring

2^x

from the denominator yields

2^x(4 + 2 + 1) = 7 \cdot 2^x

. Dividing the numerical coefficients gives

\frac{21}{7} = 3

. Substituting

4^{x-1} = 2^{2x-2}

into the ratio gives

3 \cdot \frac{2^{2x-2}}{2^x} = 3 \cdot 2^{x-2}

. Setting

3 \cdot 2^{x-2} = 24

leads to

2^{x-2} = 8 = 2^3

, which gives

x - 2 = 3

and therefore

x = 5

Factor out common terms from the numerator and denominator

Numerator:

4^{x-1}(4^2 + 4^1 + 1) = 21 \cdot 4^{x-1}

. Denominator:

2^x(2^2 + 2^1 + 1) = 7 \cdot 2^x

Grouping power terms simplifies sums of exponential expressions.

Divide the numerator by the denominator and convert bases

\frac{21 \cdot 4^{x-1}}{7 \cdot 2^x} = 3 \cdot \frac{(2^2)^{x-1}}{2^x} = 3 \cdot \frac{2^{2x-2}}{2^x}

Simplifying

\frac{21}{7} = 3

and expressing base 4 in base 2 allows applying laws of indices.

Apply the quotient rule of indices:

\frac{a^m}{a^n} = a^{m-n}

3 \cdot 2^{(2x-2) - x} = 3 \cdot 2^{x-2}

Subtracting exponents of like bases simplifies the fractional index expression.

Set the simplified expression equal to 24 and solve for

x

3 \cdot 2^{x-2} = 24 \implies 2^{x-2} = 8 = 2^3 \implies x - 2 = 3 \implies x = 5

Equating the exponents when bases are identical yields the linear equation

x - 2 = 3

Factoring sums of exponential terms and applying the quotient rule of indices

Question