Math, asked by DeFalt, 1 year ago

Maths 50 Marks
Please provide steps.

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Answered by isuru96
1
First of all Im very sorry for my mistake in the previous answer. I Corrected it ..but Im deeply  sorry and apologize for guiding you towards a wring aproach


Hi..there are many methods to approach a question like this..but aince you havent mention any specific approach , I will use logarithems and their properties to solve this one..i hope you have good understanding on the basics of properties of logarithems

Now..first of all lets use a substitution
(5 + 2√6)^(x^2 -8)   = a
(5 + 2√6)^(x^2 -8)  = b

Then  we can simplify our equation as

a  + b = 10     -----(1)

Now using the definition if logarithems ..we can write

log_ (5 + 2√6)    a  = x^2 -8       -----(2)
log _ (5 - 2√6)    b = x^2 -8       -----(3)

[ i.e , if a^b = x..the  log_a x = b ]

Now

(2) × (3)
[ log_ (5 + 2√6)    a  ] × [ log_ (5  - 2√6)    b  ]
 = (x^2 -8)^2

Now..i hope you know that
log_a b  = log_c b / log_c a

Using that property we can arrive at ..

[lg a / lg  (5 + 2√6)  ] [ lg b / lg  (5 - 2√6) ] = (x^2 -8)^2

Which means

( lg a × lg b)  / [lg(5 + 2√6)× (5 - 2√6) ] = (x^2 -8)^2

Now lets use the property

log a  × log b = log (a+b)

[lg (a +b) ] / [ lg  (5 + 2√6 + 5 - 2√6) ]  = (x^2 -8)^2

Now from (1) we know  that a+b =10..using that we  can simplify the l.h.s  to be equal to 1 !!!

 ( I leave that simplification up to you..but if you need any help there..please let me know :) )

now the things are really simple
(x^2 -8)^2 = 1

this is a polynomial of 4th degree..so this should have 4 solutions

if we solve this one...
(x^2 -8) = ±1

considering +1
×^2 - 8 = 1
x^2 = 9
x = ± 3

considering -1
x^2 -8 = -1
x^2 =7
x =±√7

now we have all 4 answers !!
i.e :
x = 3 , x= -3 , x =√7 , x = -√7

now the 2nd part should be very easy

for the 3rd part..it says if
-3<x≤ 5
what would be the value of x..

since we know all the possible values if x..you should be able to find the answer easily :)

hope this will help you..please comment if you need any help :)

DeFalt: but the property of log (a-b) : log a/ log b seems incorrect
DeFalt: can you solve this by factorising the expression?
isuru96: im sorry i have made a mistake in my first answer..i corrected it..and i will try to solve it using factorising the expression :)
isuru96: but..although i used logarithems..what ive done is using a subsitution in order to make factorization easy.. let me see if theres any direct method to use factorization :D
isuru96: I think..this might be the way you are looking in to :)
Consider
 
5^2 -(2 √6)^ 2  = ( 5 + 2 √6)( 5 - 2 √6)
25  - 24             = ( 5 + 2 √6)( 5 - 2 √6)
1                        = ( 5 + 2 √6)( 5 - 2 √6)

Now
( 5 + 2 √6)  = 1/ ( 5 - 2 √6)    -------(1)

We can substitute it in the 1st equation and get

[ 1/ ( 5 - 2 √6)]^(x^2 -8)  + ( 5 - 2 √6)^(x^2 -8) = 10

Lets substitute
      ( 5 - 2 √6)^(x^2 -8)  = t

1/t  + t =10
t^2  -10t  +1 =0
isuru96: hmmm...it seems posting an answer in the comment section doesnt work...i found the way to solve your problem.using factorization
isuru96: but i cant post it or send it as a meesage since the answer is a little bit long
isuru96: if you can repost the question..i might be able to post it..or shall i erase my answer and replace it with the second method ?
DeFalt: THANKS :))))))
DeFalt: no prob i can understand this one.
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