Find the value of x that satisfies the exponential equation \frac{100^{x+1}}{1000^{x-1}} =… — JAMB UTME

Find the value of

x

that satisfies the exponential equation

\frac{100^{x+1}}{1000^{x-1}} = 10^{2x-1}

Answer: 2

The value of

x

that satisfies the equation is 2.

Rewriting the terms in base 10 gives

\frac{10^{2x+2}}{10^{3x-3}} = 10^{2x-1}

. Using the division rule of indices yields

10^{-x+5} = 10^{2x-1}

. Equating exponents gives

-x+5 = 2x-1

, which simplifies to

3x = 6

, so

x = 2

Convert each power to base 10

100^{x+1} = (10^2)^{x+1} = 10^{2x+2}

and

1000^{x-1} = (10^3)^{x-1} = 10^{3x-3}

Converting all non-prime composite bases to powers of a common fundamental base allows exponent comparison.

Apply the quotient rule of indices to the left side

\frac{10^{2x+2}}{10^{3x-3}} = 10^{(2x+2)-(3x-3)} = 10^{-x+5}

According to the index quotient law

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

, subtract the denominator's exponent from the numerator's exponent.

Equate exponents of equal bases

-x + 5 = 2x - 1

a^f(x) = a^g(x)

for

a > 0, a \neq 1

, then

f(x) = g(x)

Solve the linear equation for

x

3x = 6 \implies x = 2

Isolate the variable

x

to find its value.

Exponential equations solvable by converting to a common base

Question