Question 30273: Can you help me solve the following:
1/2[5^x - 5^(-x)] = 3
5^(-x) is 5 raised to the power minus x
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 1/2[5^x - 5^(-x)] = 3
1/2[5^x - 5^(-x)] = 3 ----(1)
Multiplying by 2 and using a^(-m) = 1/(a^m)
[5^x-1/(5^x]) = 6
Putting 5^x = t
t-1/t = 6
Multiplying through out by t
t^2-1 = 6t
t^2-6t=1
(the LHS is of the a^2-2ab where a = t and b = 3 and so we require b^2 = 9
to make it a perfect square. and hence adding to 9 to both the sides)
t^2-6t +9 =1+9
(t-3)^2 = 10
Taking the sqroot
(t-3)= +(sqrt10) or (t-3)= -(sqrt10)
t = 3+(sqrt10) and t = 3-(sqrt10)
5^x = 3+(sqrt10) and 5^x= 3-(sqrt10)
x = log(base5)[3+(sqrt10)]----(*) and x =log(base5)[3-(sqrt10)] ----(**)
using definition b^p = N implies log(N)to the base b is equal to p
Note: But observe in x =log(base5)[3-(sqrt10)]---- (**)
that sqrt(10) >sqrt(9)
That is sqrt(10) >3
and therefore [3-(sqrt10)]=(3-something more than 3) and therefore negative and log(N) for any base is defined only for N>0
Hence this second value for x namely (**)is not defined
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Verification: x = log(base5)[3+(sqrt10)] in (1)
That is 5^x = [3+(sqrt10)] in (1) implies
LHS =(1/2){[3+(sqrt10)]-1/[3+(sqrt10)]}
=(1/2){[3+(sqrt10)]^2-1}/[3+(sqrt10)]
=(1/2)[9+10+6(sqrt10)-1]/[3+(sqrt10)]
=(1/2)[18+6(sqrt10)]/[3+(sqrt10)]
=(1/2)X6[[3+(sqrt10)]/[3+(sqrt10)]
=3 (cancelling 2 and [3+(sqrt10)] )
=RHS
Therefore our value for x given by x = log(base5)[3+(sqrt10)]is correct
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Note: OR you may take log on both the sides and this implies
log5^x = log[3+(sqrt10)] and log5^x=log[ 3-(sqrt10)]
x log5 =log[3+(sqrt10)] and x log5 =log[3-(sqrt10)]
using logbase 10 [(m)^n]= ntimesslog(m) for base 10
x =[log[3+(sqrt10)] /log5]and
x =[log[3-(sqrt10)] /log5] (the base is 10)
giving
x = log(base5)[3+(sqrt10)] ----(*) and x =log(base5)[3-(sqrt10)] ----(**)
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using log(base b) (a) =[ log(base c)(a)]/[log (base c)(b)]
{ in the above a= [3+(sqrt10)] , b = 5 and c= 10 inthe first value of x
and similarly for the other )
Note: You may directly arrive at (*) and (**) using definition
b^p = N implies log(N)to the base b is equal to p
Note: If the second value (**) is not defined
how does it hold in the equatlion
1/2[5^x - 5^(-x)] = 3 ----(1) ?
It holds because (1) is a quadratic in 5^x =t and any quadratic equation in t has to have two values and hence the other value (**) holds in the equation but is not defined in our context as log(of a negative quantity) is not defined.
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