the number of solutions of the equation 5^x+5^-x=log25/log10 (i.e log 25 to the base 10) is?

Question

the answer given in a book for this is zero. if so then how?plz explain

Answer 1

We can use log25 because the base 10 is implied (being the common log)

If you multiply the equation by 5^x, you get:

5^(2x) + 1 = log25 (5^x)

Let 5^x = p, thus

p² - log25*p + 1 = 0

This is now in quadratic form. The discriminant (b²-4ac) of a quadratic tells you the number and type of solutions. In this case, b²-4ac = (log25)² - 4 which is negative. Therefore the quadratic equation has no real solutions.

Answer 2

The equation hasn't any solution indeed.
First, you should observe that 5^(-x)=1/(5^x), according to negative exponent properties.
That way, you get 5^x + 1/(5^x) = lg 25. (where lg 25 = log25/log10, written shortly).
Note t = 5^x. Resolve the equation in t variable:
t + 1/t = lg 25. Multiply with t (obviously t cannot be zero).
Equation comes to t^2 - lg25*t +1 = 0. That is a second degree equation. The discriminant of equation is negative:
(lg 25)^2 - 4 < 0. (lg 25 < 1.4, approx.) 1.4^2 = 1.96 < 2 or 1,96^2 < 2^2 = 4.
So, (lg 25)^2 < 4. Therefore, lg 25 - 4 < 0.
With a negative discriminant, the equation in t variable has just complex solutions, but not real.
Return to notation for t. 5^x = t. So, there is no real number x that satisfy this equation. (q.e.d)

Answer 3

5^x + 5^-x must be at least 2. When x=0 it is 5^0 + 5^0 = 1 + 1 = 2.

The sum of those terms increases as x moves away from zero in either the positive or negative direction. For example, when x is 0.2 or -0.2 the sum is 2.1045, and when x is 1 or -1 the sum is 5.2.

log(base10) of 25 is less than 2... because 25 is less than 10^2, which is 100. It's about 1.39794, but you don't have to know exactly what it is to know it must be less than 2.

Since log25/log10 is less than 2, and every value of x yields a sum greater than 2, there zero solutions.

Answer 4

Log10 = 1 which means 10^1 = 10

THEREFORE THE PROBLEM BECOMES
5^x + 5^-x = log25

5^x + 5^-x = 1.397

then xlog5 -xlog5 = 1.397

but xlog5-xlo5 = 0 therefor the equation is indeterminant, or there are zero solutions.