Carcass wrote:
If \(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\) then x must equal which of the following ?
Indicate
all that apply.
❑ 2a
❑ 2b
❑ 7a - 2b
❑ \(\frac{a}{7}\)
❑ \(\frac{b}{2}\)
Here given
\(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\)
\(3^{2a}\)\(11^b\) = \(3^{12x}\) \(3^{2x}\) \(11^{2x}\) (Since \(27^{4x}\) = \({3^{(3x)}}^{4}\) = \(3^{12x}\))
or \(3^{2a}\)\(11^b\) = \(3^{14x}\) \(11^{2x}\)
Since prime bases are same, the exponents must also be equal.
14x = 2a,
or x= \(\frac{2}{14}\)
or a =\(\frac{a}{7}\)
And 2x = b, or x= \(\frac{b}{2}\)
Therefore only choices (D) and (E) must be true