Multiply the following: 7x^2√2x^8 * 6x^4√23x^9 * 3x^16
564159456 sqrt(23) x^(61/2)
i do not have that
\[7x^2 * \sqrt{2} x^4*6x^4*\sqrt{23}*x^{9/2}*3x^{16}\]Remember that when exponents of the same base are multiplied, they exponents add. The square root has an exponent of 1/2.
hm
(54 x^22) 2 x^7 = 108 x^29
Therefore, we would have \[(7*\sqrt{2}*6*\sqrt{24} * 3)*x^{2+4+(9/2) + 16}\]
i have like 126x^30
kayser, let me verify that equation you typed. Is it \[7x^2*\sqrt{2x^8}*6x^4*\sqrt{23x^9}*3x^{16}\]
sir the 6x^4 is with √23x^9
it is like this 6x^4√23x^9
\[6x^{4\sqrt{23x^9}}\]??
yes
what would it be then
Give me a second to type it up.
ok
\[7x^{2*\sqrt{2}*x^4}*6x^{4\sqrt{23} \sqrt{x^9}}*3x^{16}\]This further simplifies to \[(7*6*3)^{(2*\sqrt{2}*x^4 + 4*\sqrt{23}x^{9/2}+16)}\] I don't think we can go any further in the exponent.
kayser, please see my updated solution to the inequality question you asked earlier.
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