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Please solve for x. 2log5 + 3logx = 2

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damdoomdima | Student, Grade 11 | (Level 2) Honors

Posted August 4, 2010 at 8:15 AM via web

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Please solve for x.

2log5 + 3logx = 2

4 Answers | Add Yours

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 4, 2010 at 8:17 AM (Answer #1)

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2 log 5 + 3 log x = 2

==> log 5^2 + log x^3 = 2

We know that: log a + log b = log a*b

==> log 25*x^3 = 2

==> 25x^3 = 10^2

==> 25x^3 = 100

Divide by 25:

==> x^3 = 100/25 = 4

==> x = (4)^1/3

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted August 4, 2010 at 2:55 PM (Answer #2)

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First, we'll use the power property of logarithms, for the terms of the expression:

2 log 5 = log 5^2

3 log x = log x^3

2 log 5 + 3 log x  = log 5^2 +  log x^3

Now, we'll use the product property of logarithms:

log 5^2 +  log x^3 = log 25*x^3

log 25*x^3 = 2

25*x^3 = 10^2

25*x^3 = 100

We'll divide by 25 both sides:

x^3 = 4

x = 4^(1/3)

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neela | High School Teacher | (Level 3) Valedictorian

Posted August 4, 2010 at 4:56 PM (Answer #3)

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To solve for x in 2log5 +3logx =2

Solution:

2log5 = log 5^2 =log25, as  n*log b = logb^n

3logx = logx^3

and 2 = lof10^2 = log 100.

Replacin in  the given equation:

log25*logx^3 = log 100.

log(25x^3) = log 100. Taking antilogarithms,

25x^3 = 100

x^3 = 100/25 = 4

x^3 = 4. Take the cube root.

x = 4^(1/3) is the real solution.

 

 

 

 

 

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yamaguchityler | Student, Undergraduate | (Level 2) Assistant Educator

Posted January 31, 2014 at 12:16 AM (Answer #4)

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To solve, do the following steps:

2log5 + 3logx = 2

-> log5^2 + logx^3 = 2

-> = log25*x^3 = 2

-> 25*x^3 = 10^2

-> 25*x^3 = 100

-> Divide by 25

-> x^3 = 4

-> Take the cube root

-> x = (4)^(1/3)

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