now, let's take a peek at the denominators, we have 3 and 8 and 12, we can get an LCD of 24 from that.
Let's multiply both sides by the LCD of 24, to do away with the denominators.
so, let's recall that a whole is "1", namely 500/500 = 1 = whole, or 5/5 = 1 = whole or 24/24 = 1 = whole. So the whole class will yield a fraction of 1/1 or just 1.
![\bf ~\hspace{7em}\stackrel{\textit{basketball}}{\cfrac{1}{3}}+\stackrel{\textit{soccer}}{\cfrac{1}{8}}+\stackrel{\textit{football}}{\cfrac{5}{12}}+\stackrel{\textit{baseball}}{x}~=~\stackrel{\textit{whole}}{1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{24}}{24\left(\cfrac{1}{3}+\cfrac{1}{8}+\cfrac{5}{12}+x \right)=24(1)}\implies (8)1+(3)1+(2)5+(24)x=24 \\\\\\ 8+3+10+24x=24\implies 21+24x=24\implies 24x=3 \\\\\\ x=\cfrac{3}{24}\implies x=\cfrac{1}{8}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B7em%7D%5Cstackrel%7B%5Ctextit%7Bbasketball%7D%7D%7B%5Ccfrac%7B1%7D%7B3%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsoccer%7D%7D%7B%5Ccfrac%7B1%7D%7B8%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bfootball%7D%7D%7B%5Ccfrac%7B5%7D%7B12%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bbaseball%7D%7D%7Bx%7D~%3D~%5Cstackrel%7B%5Ctextit%7Bwhole%7D%7D%7B1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B24%7D%7D%7B24%5Cleft%28%5Ccfrac%7B1%7D%7B3%7D%2B%5Ccfrac%7B1%7D%7B8%7D%2B%5Ccfrac%7B5%7D%7B12%7D%2Bx%20%5Cright%29%3D24%281%29%7D%5Cimplies%20%288%291%2B%283%291%2B%282%295%2B%2824%29x%3D24%20%5C%5C%5C%5C%5C%5C%208%2B3%2B10%2B24x%3D24%5Cimplies%2021%2B24x%3D24%5Cimplies%2024x%3D3%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B3%7D%7B24%7D%5Cimplies%20x%3D%5Ccfrac%7B1%7D%7B8%7D)
Step-by-step explanation:
1. 
multiply

do both the numerator and denominater

Answer:
100 years old
Step-by-step explanation:hope it helps:)
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Step-by-step explanation:
Let us revise some logarithmic rules
∵ 10 log(x) + 5 log(64)
- At first re-write 10 log(x)
∴ 10 log(x) = 
- Then re-write 5 log(64)
∴ 5 log(64) = 
∴ 10 log(x) + 5 log(64) =
+ 
- Use the 3rd rule above to make it single logarithm
∵
+
= ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∵ 64 = 2 × 2 × 2 × 2 × 2 × 2
∴ We can write 64 as 
∴ 
- Multiply the two powers of 2
∴ 
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Learn more:
You can learn more about the logarithmic functions in brainly.com/question/11921476
#LearnwithBrainly
Answer:
Step-by-step explanation:
www.564bdyjkthl