Question

what values of a and b make the equation true?\\(\sqrt{648} = \sqrt{2^a \cdot 3^b}\\)\
\\(a=4, b=3\\)\
\\(a=3, b=4\\)\
\\(a=2, b=3\\)\
\\(a=3, b=2\\)

Explanation:

Step1: Factorize 648

First, we factorize 648. We know that \(648 = 2\times324=2\times2\times162 = 2\times2\times2\times81=2^{3}\times81\). And \(81 = 3^{4}\), so \(648=2^{3}\times3^{4}\)? Wait, no, wait. Wait, \(2\times2\times2 = 8\), \(8\times81 = 648\). Wait, \(81 = 3^4\)? No, \(3^4=81\)? Wait, \(3^3 = 27\), \(3^4=81\), yes. Wait, but let's do prime factorization properly. \(648\div2 = 324\), \(324\div2 = 162\), \(162\div2 = 81\), \(81\div3 = 27\), \(27\div3 = 9\), \(9\div3 = 3\), \(3\div3 = 1\). So the prime factors are \(2^3\times3^4\)? Wait, no, we divided by 2 three times and by 3 four times? Wait, \(2\times2\times2 = 8\), \(3\times3\times3\times3 = 81\), \(8\times81 = 648\). So \(648 = 2^{3}\times3^{4}\)? Wait, but the square root of 648 is \(\sqrt{648}=\sqrt{2^{3}\times3^{4}}\). Wait, but the problem says \(\sqrt{648}=\sqrt{2^{a}\times3^{b}}\). Wait, no, the problem is \(\sqrt{648}=\sqrt{2^{a}\cdot3^{b}}\)? Wait, no, the original equation is \(\sqrt{648}=\sqrt{2^{a}\cdot3^{b}}\)? Wait, no, looking back, the equation is \(\sqrt{648}=\sqrt{2^{a}\cdot3^{b}}\)? Wait, no, the user's image shows \(\sqrt{648}=\sqrt{2^{a}\cdot3^{b}}\)? Wait, no, maybe it's \(\sqrt{648}=2^{a}\cdot3^{b}\)? Wait, the image says "√648 = √2ᵃ·3ᵇ"? Wait, no, maybe a typo. Wait, the options are for a and b. Let's check the options. Let's compute \(\sqrt{648}\). First, simplify \(\sqrt{648}\). \(648 = 8\times81 = 8\times9^2\), so \(\sqrt{648}=\sqrt{8\times81}=\sqrt{8}\times9 = 3\sqrt{8}=3\times2\sqrt{2}=6\sqrt{2}\)? Wait, no, that's wrong. Wait, \(81\) is \(9^2\), so \(\sqrt{648}=\sqrt{8\times81}=\sqrt{8}\times\sqrt{81}=3\sqrt{8}=3\times2\sqrt{2}=6\sqrt{2}\)? No, that's not right. Wait, \(8 = 2^3\), so \(\sqrt{8}=2^{3/2}\), so \(\sqrt{648}=9\times2^{3/2}=3^2\times2^{3/2}\). But the right-hand side is \(\sqrt{2^a\cdot3^b}\)? Wait, no, maybe the equation is \(\sqrt{648}=2^{a}\cdot3^{b}\). Let's check the options. Let's compute \(2^a\cdot3^b\) for each option.

Option 1: \(a = 4, b = 3\): \(2^4\times3^3=16\times27 = 432\). \(\sqrt{648}\approx25.45\), 432 is not equal to that. Wait, no, maybe the equation is \(\sqrt{648}=2^{a}\cdot3^{b}\). Wait, let's compute \(\sqrt{648}\). \(648 = 2\times324 = 2\times2\times162 = 2\times2\times2\times81 = 2^3\times3^4\). So \(\sqrt{648}=\sqrt{2^3\times3^4}=\sqrt{2^2\times2\times3^4}=2\times3^2\times\sqrt{2}=9\times2\times\sqrt{2}=18\sqrt{2}\approx25.45\). Wait, but let's check the options. Wait, maybe the equation is \(648 = 2^a\times3^b\) (without the square root). Let's check that. If \(648 = 2^a\times3^b\), then from prime factorization, we have \(2^3\times3^4\), so \(a = 3\), \(b = 4\)? But that's not an option. Wait, the options are:

1. \(a = 4, b = 3\)

2. \(a = 3, b = 2\)

3. \(a = 3, b = 4\)

4. \(a = 2, b = 3\)

Wait, let's compute \(2^a\times3^b\) for each option:

- Option 1: \(2^4\times3^3 = 16\times27 = 432 eq648\)

- Option 2: \(2^3\times3^2 = 8\times9 = 72 eq648\)

- Option 3: \(2^3\times3^4 = 8\times81 = 648\). Yes! So that's correct. Wait, but the equation in the image is \(\sqrt{648}=\sqrt{2^a\cdot3^b}\)? Wait, maybe a misprint. If the equation is \(648 = 2^a\times3^b\), then \(a = 3\), \(b = 4\) is correct. But let's check the square root. If the equation is \(\sqrt{648}=2^a\times3^b\), then \(\sqrt{648}\approx25.45\). Let's compute \(2^a\times3^b\) for option 3: \(2^3\times3^4 = 8\times81 = 648\), which is not the square root. Wait, maybe the equation is \(\sqrt{648}=2^a\times3^b\). Then \(\sqrt{648}\approx25.45\). Let's compute option 4: \(a = 2, b = 3\): \(2^2\times3^3 = 4\times27 = 108 eq25.45\). Option 1: \(16\times27 = 432\). Option 2: \(8\times9 = 72\). Option 3: \(8\times81 = 648\). Wait, maybe the equation is \(648 = 2^a\times3^b\), so \(a = 3\), \(b = 4\) is correct. But let's check the prime factorization again. \(648\div2 = 324\), \(324\div2 = 162\), \(162\div2 = 81\) (so three 2s), then \(81\div3 = 27\), \(27\div3 = 9\), \(9\div3 = 3\), \(3\div3 = 1\) (four 3s). So \(648 = 2^3\times3^4\), so \(a = 3\), \(b = 4\). But looking at the options, one of them is \(a = 3, b = 4\)? Wait, the third option (middle bottom) is \(a = 3, b = 4\)? Wait, the image shows:

Top right: \(a = 4, b = 3\)

Top left: \(a = 3, b = 2\)

Middle bottom: \(a = 3, b = 4\)

Bottom: \(a = 2, b = 3\)

So the correct one is \(a = 3, b = 4\), which is the middle bottom option. Wait, but let's verify. If we take \(2^3\times3^4 = 8\times81 = 648\), which is correct. So the values of \(a = 3\) and \(b = 4\) make the equation \(648 = 2^a\times3^b\) true (assuming the square root was a typo). So the correct option is the one with \(a = 3, b = 4\).

Step2: Verify each option

- Option \(a = 4, b = 3\): \(2^4\times3^3 = 16\times27 = 432 eq 648\)
- Option \(a = 3, b = 2\): \(2^3\times3^2 = 8\times9 = 72 eq 648\)
- Option \(a = 3, b = 4\): \(2^3\times3^4 = 8\times81 = 648\) (matches)
- Option \(a = 2, b = 3\): \(2^2\times3^3 = 4\times27 = 108 eq 648\)

Answer:

\(a = 3, b = 4\) (the option with \(a = 3, b = 4\))