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balu736 [363]
3 years ago
13

How do you solve this equation?

Mathematics
1 answer:
Travka [436]3 years ago
7 0
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


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vivado [14]

Answer:  this is our required factor i.e.

8z^3+27=(2z+3)(4z^2-6z+9)

Explanation:

Since we have given that

8z^3+27

As we know the identity , which says that

a^3+b^3=(a+b)(a^2-ab+b^2)

So, we can use this here ,

8z^3+27=(2z)^3+3^3\\\\=(2z+3)((2z)^2-2z\times 3+3^2)\\\\=(2z+3)(4z^2-6z+9)

Hence this is our required factor i.e.

8z^3+27=(2z+3)(4z^2-6z+9)


7 0
3 years ago
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