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Damm [24]
4 years ago
6

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

Mathematics
2 answers:
zhuklara [117]4 years ago
4 0

Answer:

Yes mark me brainliest

Step-by-step explanation:


marshall27 [118]4 years ago
3 0

Answer:

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


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