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Indices and Laws of Indices

22 questions

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Question 21Question →

If (1681)x−1×(278)x+2=94\left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{4}, find the value of xx.

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Answer: 88

Answer

The value of xx is 88.
By writing 1681\frac{16}{81} as (23)4\left(\frac{2}{3}\right)^4, 278\frac{27}{8} as (23)−3\left(\frac{2}{3}\right)^{-3}, and 94\frac{9}{4} as (23)−2\left(\frac{2}{3}\right)^{-2}, the equation simplifies via exponent addition to (23)4x−4−3x−6=(23)−2\left(\frac{2}{3}\right)^{4x - 4 - 3x - 6} = \left(\frac{2}{3}\right)^{-2}. Equating indices gives x−10=−2x - 10 = -2, which yields the correct solution x=8x = 8.

Step-by-Step Solution

1
Express all fractional terms with a common base of 23\frac{2}{3}
1681=(23)4\frac{16}{81} = \left(\frac{2}{3}\right)^4, 278=(32)3=(23)−3\frac{27}{8} = \left(\frac{3}{2}\right)^3 = \left(\frac{2}{3}\right)^{-3}, and 94=(32)2=(23)−2\frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}
Converting all terms to a single common base allows exponents to be combined using the laws of indices.
2
Substitute the common base expressions back into the original equation
((23)4)x−1×((23)−3)x+2=(23)−2\left( \left(\frac{2}{3}\right)^4 \right)^{x-1} \times \left( \left(\frac{2}{3}\right)^{-3} \right)^{x+2} = \left(\frac{2}{3}\right)^{-2}
Applying the power of a power law (am)n=amn(a^m)^n = a^{mn} to simplify each term.
3
Apply the power law and multiplication law of indices
(23)4(x−1)×(23)−3(x+2)=(23)−2  ⟹  (23)4(x−1)−3(x+2)=(23)−2\left(\frac{2}{3}\right)^{4(x-1)} \times \left(\frac{2}{3}\right)^{-3(x+2)} = \left(\frac{2}{3}\right)^{-2} \implies \left(\frac{2}{3}\right)^{4(x-1) - 3(x+2)} = \left(\frac{2}{3}\right)^{-2}
When multiplying exponential terms with identical bases, add their exponents: am×an=am+na^m \times a^n = a^{m+n}.
4
Equate the exponents and solve for xx
4(x−1)−3(x+2)=−2  ⟹  4x−4−3x−6=−2  ⟹  x−10=−2  ⟹  x=84(x-1) - 3(x+2) = -2 \implies 4x - 4 - 3x - 6 = -2 \implies x - 10 = -2 \implies x = 8
Since the bases on both sides are equal and non-zero, their indices must be equal.

Key Concept

Laws of Indices: Base Conversion and Exponential Equations
Estimated Time:2m 0s
Question 22Question →

If 52x−1×25x+1=125x+25^{2x - 1} \times 25^{x + 1} = 125^{x + 2}, what is the value of xx?

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Answer: 5

Answer

The value of xx is 55.
Converting all terms to base 55 gives 52x−1×52(x+1)=53(x+2)5^{2x - 1} \times 5^{2(x + 1)} = 5^{3(x + 2)}. Simplifying the exponents yields 52x−1+2x+2=53x+65^{2x - 1 + 2x + 2} = 5^{3x + 6}, which reduces to 54x+1=53x+65^{4x + 1} = 5^{3x + 6}. Setting the exponents equal to each other gives 4x+1=3x+64x + 1 = 3x + 6, resulting in x=5x = 5.

Step-by-Step Solution

1
Express all bases in terms of prime base 5
52x−1×(52)x+1=(53)x+25^{2x - 1} \times (5^2)^{x + 1} = (5^3)^{x + 2}
All terms must share the same base to combine exponents using index laws.
2
Apply power of a power rule (am)n=amn(a^m)^n = a^{mn}
52x−1×52x+2=53x+65^{2x - 1} \times 5^{2x + 2} = 5^{3x + 6}
Multiplication of inner and outer powers simplifies composite exponent expressions.
3
Apply the product rule am×an=am+na^m \times a^n = a^{m+n}
54x+1=53x+65^{4x + 1} = 5^{3x + 6}
Adding the exponents on the left-hand side produces a single exponential term.
4
Equate exponents and solve for xx
x=5x = 5
Equal bases imply equal exponents: 4x+1=3x+64x + 1 = 3x + 6.

Key Concept

Solving exponential equations using common base conversion and laws of indices
Estimated Time:1m 30s
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