Is this the correct procedure for this kind of question? apparently the answer is supposed to be log6/log5.

log 3                        

= x                  

log 5

Step-by-step explanation:

25 ^x - 5^x +4 = 10

Subtract 4 from each side

25 ^x - 5^x +4-4 = 10-4

25 ^x - 5^x  = 6

25 can we rewritten as 5^2

5^2^x - 5^x  = 6

a^b^c can be written as a^(b*c)

5^(2x) - 5^x  = 6

Replace 5^x with m

m^2 - m = 6

Subtract 6 from each side

m^2 -m -6 = 6-6

m^2 -m-6=0

Factor

(m-3) (m+2) = 0

Using the zero product property

m-3 = 0   m+2=0

m=3      m=-2

Substitute m=x^2 back in

5^x =3             5^x =-2

Take the log of each side

log5 (3) =x         log5 (-2) =x

There is no  log -2  so this is not a solution

Assuming we want to use base 10

log 3                        

= x                  

log 5

answer: the answer is 35%

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !