Question

Iog(2)24-½log(2)72_______________ =Iog(3)18-⅓log(3)72-​

Answer 1

log(2) 24 - 1/2*log(2) 72 =log(2) 24 - log(2) √72 = log(2) 24 - log(2) √(36*2) = log(2) 24/√(36*2)  = log(2) 24/6√2) = log(2) 4/√2 = log(2) 4 - log(2) √2 = 2-1/2 = 3/2

log(3) 18 - 1/3log(3) 72 =log(3) 18 - log(3) ∛72 = log(3) 18 - log(3) ∛(8*9) = log(3) 18/∛(2³*9)  = log(3) 18/2∛9 = log(3) 9/∛9 = log(3)9 - log(3) ∛9 = 3 - 2/3 = 6/3 - 2/3 = 4/3

3/2 : 4/3 = 3/2 * 3/4 = 9/8

Iog(2)24-½log(2)72         9

_______________ =     -----

Iog(3)18-⅓log(3)72           8

Answer 2

Ответ:

Прикрепила фото

Объяснение:

Answer 3

It seems like you have an equation involving logarithms that you'd like to solve. The equation you provided is:

$\log_2(24) - \frac{1}{2}\log_2(72) = \log_3(18) - \frac{1}{3}\log_3(72)$

Let's break down the equation and solve for the unknown variable.

1. Simplify the logarithmic expressions:

   $\log_2(24) = \log_2(8 \cdot 3) = \log_2(8) + \log_2(3) = 3 + \log_2(3)$ $\log_2(72) = \log_2(8 \cdot 9) = \log_2(8) + \log_2(9) = 3 + 2\log_2(3)$ $\log_3(18) = \log_3(9 \cdot 2) = \log_3(9) + \log_3(2) = 2 + \log_3(2)$ $\log_3(72) = \log_3(9 \cdot 8) = \log_3(9) + \log_3(8) = 2 + \log_3(8)$

2. Substitute these simplified expressions back into the original equation:

   $(3 + \log_2(3)) - \frac{1}{2}(3 + 2\log_2(3)) = (2 + \log_3(2)) - \frac{1}{3}(2 + \log_3(8))$

3. Distribute the coefficients:

   $3 + \log_2(3) - \frac{3}{2} - \log_2(3) = 2 + \log_3(2) - \frac{2}{3} - \frac{1}{3}\log_3(8)$

4. Simplify:

   $\frac{1}{2} = \frac{5}{3} - \frac{1}{3}\log_3(8)$

5. Solve for $\log_3(8)$:

   $\frac{1}{3}\log_3(8) = \frac{5}{3} - \frac{1}{2} = \frac{4}{6}$ $\log_3(8) = 4$

So, the solution to the equation is:

$\log_3(8) = 4$

If you have any other equations or questions, feel free to ask!