Carcass wrote:
\(b>1\)
\(\frac{(b^4)^x (b^{30})}{(b^2)^7} < (b^x)(b^{37})\)
Quantity A |
Quantity B |
\(x\) |
\(8\) |
A few important exponent rules:
(1) \((n^x)^y = n^{xy}\)
(2) \((n^x)/(n^y) = n^{x-y}\)
(3)\((n^x)(n^y) = n^{x+y}\)
(4) If \(n^x = n^y\), then \(x=y\)
Now, on to the problem:
The denominator on the left side can be simplified to \(b^{14}\).
\(\frac{b^{30}}{b^{14}} = b^{16}\).
Then, simplify the other term in the numerator to \(b^{4x}\).
Multiply that by \(b^{16}\) and we are left with \(b^{4x+16}\) on the left side of the equation.
The other side simplifies to \(b^{x+37}\). Since the bases are the same, using rule 4 we can create a new equation: 4x+16 < x+37. Solve for x and we we get x<7.
Since x is less than 7, Quantity B will always be greater than Quantity A. Thus the correct answer is B.