# What is the real number m if 49^x+m*7^x-m-1=0 ?

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We'll re-write the equation, replacing 7^x by t;

t^2 + m*t - (m+1) = 0

We'll consider the cases:

1) delta > 0 and the product of roots P<0.

2) delta = 0 and the sum of roots S>0.

delta = b^2 - 4ac

a = 1, b = m and c = -(m+1)

delta = m^2 + 4(m+1)

delta >=0 => m^2 + 4m + 4 = (m+2)^2 > 0

We'll impose the constraint P<0

P = x1*x2 = -(m + 1)

-(m + 1) < 0

m + 1> 0

m>-1

We'll impose the constraint S>0.

m>0

We'll impose the constraint delta = 0 => (m+2)^2 => m = -2.

**The intervals of real values of m, such as 49^x+m*7^x-m-1=0 are: {-2}U(0 , +infinite).**

from t^2 + m*t -m - 1 = 0

m(t -1) + t^2 - 1 = 0

if t is not equal to 1, cancel t -1, we have

m = -( t+ 1) = - (7^x + 1). for x is not equal to 0.

if x is equal to 0, m can be any real value.