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Show that 32x1=9x33^{2x - 1} = \frac{9^x}{3}

Full solution

Q. Show that 32x1=9x33^{2x - 1} = \frac{9^x}{3}
  1. Simplify Right Side: Let's start by simplifying the right side of the equation, (9x)/3(9^x)/3. We know that 99 is 323^2, so we can rewrite 9x9^x as (32)x(3^2)^x.
  2. Apply Exponent Property: Now, using the property of exponents that (ab)c=abc(a^b)^c = a^{b*c}, we can simplify (32)x(3^2)^x to 32x3^{2*x}. So, 9x3\frac{9^x}{3} becomes 32x3\frac{3^{2*x}}{3}.
  3. Use Exponent Rule: Next, we can use the property of exponents that am/an=amna^{m}/a^{n} = a^{m-n} to simplify 32x/33^{2*x}/3. Since 33 is the same as 313^1, we can write 32x/33^{2*x}/3 as 32x13^{2*x - 1}.
  4. Finalize Equation: Now we have 32x13^{2x - 1} on both sides of the equation.\newlineThis means that 32x13^{2x - 1} is indeed equal to (9x)/3(9^x)/3.

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