Question

11. If a² + b2 = 22ab, then 2 log (a + b) =
1) log a + log b + 3log 2 + log 3
2) log a + log b + 3 log 2 + log 5
3) log a + log b + 2 log 2 + log 3
4) log a + log b + 2 log 2 + log 5​

Answer 1

Answer:

loga,logb,logc are in A.P.

⇒2logb=loga+logc

⇒logb2=log(ac)

⇒b2=ac⇒a,b,c are in G.P.

loga−log2b,log2b−log3c,log3c−loga are in A.P.

⇒2(log2b−log3c)=(loga−log2b)+(log3c−loga)

⇒3log2b=3log3c⇒2b=3c

Now, b2=ac⇒b2=a.32b⇒b=32a,c=94a

i.e., a=a,b=32a,c=94a

⇒a:b:c=1:32:94=9:6:4

Since, sum of any two is greater than the 3rd,a,b,c form a triangle

Answer 2

Answer:

mark above one as brainliest plss