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