If $9^{2 x-1}-81^{x-1}=1944$, then $x$ is

Given: $9^{2x-1} - 81^{x-1} = 1944$

First, we need to express everything in terms of base 9. Notice that $81 = 9^2$.

So: $81^{x-1} = (9^2)^{x-1}$

Using the power rule: $(a^m)^n = a^{mn}$

Therefore: $81^{x-1} = 9^{2(x-1)} = 9^{2x-2}$

Our equation becomes: $9^{2x-1} - 9^{2x-2} = 1944$

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We can factor out $9^{2x-2}$ from both terms:

$9^{2x-1} - 9^{2x-2} = 9^{2x-2} \cdot 9^1 - 9^{2x-2} \cdot 1$

$= 9^{2x-2}(9 - 1) = 9^{2x-2} \cdot 8$

So our equation is: $9^{2x-2} \cdot 8 = 1944$

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Dividing both sides by 8:

$9^{2x-2} = \dfrac{1944}{8} = 243$

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Now we need to find what power of 9 equals 243.

Let us check: $9^1 = 9$, $9^2 = 81$, $9^3 = 729$ (too big)

Since 243 is between $9^2 = 81$ and $9^3 = 729$, the exponent must be between 2 and 3.

Let us try $9^{2.5}$:

$9^{2.5} = 9^{2 + 0.5} = 9^2 \cdot 9^{0.5} = 81 \cdot \sqrt{9} = 81 \cdot 3 = 243$

Therefore: $243 = 9^{2.5}$

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Since $9^{2x-2} = 243 = 9^{2.5}$, and the bases are equal, the exponents must be equal:

$2x - 2 = 2.5$

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$2x - 2 = 2.5$

$2x = 2.5 + 2 = 4.5$

$x = \dfrac{4.5}{2} = \dfrac{9}{4}$

Answer: $x = \dfrac{9}{4}$

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Key Learning Points:

When dealing with exponential equations, always try to express everything in the same base

Factor out common exponential terms to simplify

When bases are equal, exponents must be equal

$9^{2.5} = 9^2 \cdot 9^{0.5} = 81 \cdot 3 = 243$ is a useful calculation to remember