What is the solution of the equation log15 [1/(3^x + x - 5)]=x*(log15 5 - 1) ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1).

log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1)

=> log(15) [1/(3^x + x - 5)] = x*log(15) 5 - x

=> log(15) [1/(3^x + x - 5)] = x*log(15) 5 - log (15) 15^x

=> log(15) [1/(3^x + x - 5)] = log(15) (5^x/ 15^x)

=> [1/(3^x + x - 5)] = 5^x/ 15^x

=> [1/(3^x + x - 5)] = 1/ 3^x

=> 3^x = 3^x + x - 5

=> x - 5 = 0

=> x = 5

We get x = 5

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the left term of the equation:

 log15 [1/(3^x + x - 5)] =  log15 [(3^x + x - 5)^-1]

We'll apply the power rule of logarithms:

log15 [(3^x + x - 5)^-1] = -log15 (3^x + x - 5)

We'll remove the brackets from the right side:

x*(log15 5 - 1) = xlog15 5 - x

x*(log15 5 - 1) = log15 (5^x) - x*log15 15

x*(log15 5 - 1) = log15 (5^x) - log15 15^x

Since the bases are matching, we'll apply the quotient rule:

log15 (5^x) - log15 15^x = log15 (5^x/15^x)

log15 (5^x/15^x) = log15 (5^x/5^x*3^x)

We'll simplify and we'll get;

log15 (5^x/5^x*3^x) = log15 (1/3^x)

log15 (1/3^x) =-log15 3^x

We'll re-write the equation:

 -log15 (3^x + x - 5) = -log15 3^x

We'll add log15 3^x:

log15 3^x -log15 (3^x + x - 5) = 0

Since the bases are matching, we'll apply the quotient rule:

log15[3^x/(3^x + x - 5)] = 0

[3^x/(3^x + x - 5)] = 15^0

[3^x/(3^x + x - 5)] = 1

3^x = 3^x + x - 5

We'll eliminate like terms:

0 = x - 5

x = 5

The solution of the equation is x = 5.

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