dare to solve this logarithmic problem

acc to pritishmast ans should be -0.265 but it is not in the options means that method is wrong as cal. are correct.....

qwerty can u or anyone solve this and give me the answer with ofcourse correct method........

wat i mean is , u cant apply analogy anywhere randomly, i hope u get wat i m trying to say

btw if u assumed something , then check whether the ans is within your assumption. i.e , u assumed y to be infintely large, but it actually came out to be 1-3log2 , so the final ans is contradictory to the assumption , so the assumption shud be wrng

1-3log2 is not ∞

and 2 + (1-3log2 ) ≠1-3log2

u applied a concept , do u kno the basics of dat concept ?

according to given solution ans will be same for every question like that
i.e. log₇log₇y=y
for any value of y which is an ∞ series
Ex.
y=">

how can you take log₇log₇√7+√7+√7....∞ =y
and also √7+√7+√7....∞ =y

solution....

let the infinite term be y.then

log₇log₇y

taking log on an indefinite i.e.,∞ no. gives infinite no. so,

log₇log₇y = y

differentiating both sides w.r.t. y gives

1log₇y X log₇e X 1y X log₇e=1

log7logy X logelog7 X 1y X logelog7=1

1logy.log7.y=1

y.logy=1log7

again differentiating w.r.t. y

1 + logy =0

logy=-1

y=110

y=.1

or y=1-3log2

that is the answer.....

ur steps r right but method seems to be wrong............

answer is not -0.265

oh so it is like this:
\sqrt{7+\sqrt{7+\sqrt{7+\sqrt{7+\sqrt{7}}}}..............infinity }

let y=log₇log₇(z)

where z = \sqrt{7+\sqrt{7+\sqrt{7+\sqrt{7+\sqrt{7}}}}..............infinity }

or z = \sqrt{7+z}

solving this we get z= 1±√292

z=1-√292 neglected because
log₇(1-√292 )=undefined as 1-√292 < 0

so z= 1+√292

so required value of y = log₇log₇(1+√292)

y = -0.265

hey, (7+......∞)≠(7+8+9+....∞)

dots signify that the series continues....

7+....∞={7+[7+(7+....∞)^1/2]^1/2}^1/2

now solve it..........

is the ans 1???