The value of [tex]x[/tex] is 29.
The formulae for the surface area ([tex]A_{s}[/tex]) and volume ([tex]V[/tex]) are, respectively:
[tex]A_{s} = 2\cdot (h\cdot l + w\cdot l + h\cdot w)[/tex] (1)
[tex]V = w\cdot h \cdot l[/tex] (2)
Where:
According to the statement, we must equalize (1) and (2) and substitute the remaining variables:
[tex]2\cdot (h\cdot l + w\cdot l + h\cdot w) = w\cdot h\cdot l[/tex]
[tex]2\cdot h\cdot l + 2\cdot w \cdot l + 2\cdot h\cdot w = w \cdot h \cdot l[/tex]
[tex]2\cdot \log_{2}x\cdot \log_{4}x+2\cdot \log_{3}x\cdot \log_{4}x+2\cdot \log_{2}x\cdot \log_{3}x = \log_{2}x\cdot \log_{3}x\cdot \log_{4}x[/tex]
By applying logarithm properties, we simplify the expression:
[tex]2\cdot \log_{2}x\cdot \left(\frac{\log_{2}x}{\log_{2}4} \right) + 2\cdot \left(\frac{\log_{2}x}{\log_{2}3} \right)\cdot \left(\frac{\log_{2}x}{\log_{2}4} \right) + 2\cdot \log_{2}x\cdot \left(\frac{\log_{2}x}{\log_{2}3} \right) = \log_{2}x \cdot \left(\frac{\log_{2}x}{\log_{2}3} \right)\cdot \left(\frac{\log_{2}x}{\log_{2}4} \right)[/tex]
[tex]2\cdot \left[\frac{1}{\log_{2}4}+\frac{1}{\log_{2}3\cdot \log_{2}4}+\frac{1}{\log_{2}3}\right] = \left(\frac{1}{\log_{2}3\cdot \log_{2}4} \right)\cdot \log_{2}x[/tex]
[tex]2\cdot \left(\frac{\log_{2}3+\log_{2}4+1}{\log_{2}3\cdot \log_{2}4} \right) = \left(\frac{1}{\log_{2}3 \cdot \log_{2}4} \right)\cdot \log_{2}x[/tex]
[tex]\log_{2}x = 2\cdot (\log_{2}3 + \log_{2}4+1)[/tex]
[tex]\log_{2}x = \log_{2}9 + \log_{2}16 +2[/tex]
[tex]x = 9 + 16 + 4[/tex]
[tex]x = 29[/tex]
The value of [tex]x[/tex] is 29.
We kindly invite to check this question on logarithms: https://brainly.com/question/20838017
Nota - The statement of this question presents typing mistakes and is also incomplete, the corrected form is presented below:
A right rectangular prism whose surface area and volume are numerically equal has edge lengths [tex]\log_{2} x[/tex] (height), [tex]\log_{3}x[/tex] (width) and [tex]\log_{4}x[/tex] (length). What is [tex]x[/tex]?
