Question

what value of k makes the equation true?\\((5a^{2}b^{3})(6a^{k}b)=30a^{6}b^{4}\\)\\(\bigcirc\\ 2\\)\\(\bigcirc\\ 3\\)\\(\bigcirc\\ 4\\)\\(\bigcirc\\ 8\\)

Explanation:

Step1: Multiply the coefficients and use the product rule for exponents

First, multiply the coefficients: \(5\times6 = 30\). Then, for the variables with base \(a\), use the rule \(a^m\times a^n=a^{m + n}\), so \(a^{2}\times a^{k}=a^{2 + k}\). For the variables with base \(b\), \(b^{3}\times b^{1}=b^{3+1}=b^{4}\) (since \(b = b^{1}\)). So the left - hand side of the equation becomes \(30a^{2 + k}b^{4}\).

Step2: Set up the equation for the exponents of \(a\)

We know that the right - hand side of the equation is \(30a^{6}b^{4}\). Since the coefficients and the exponents of \(b\) are already equal on both sides, we can set the exponents of \(a\) equal to each other. So we have the equation \(2 + k=6\).

Step3: Solve for \(k\)

Subtract 2 from both sides of the equation \(2 + k=6\). We get \(k=6 - 2=4\).

Answer:

4 (corresponding to the option "4")