Asked by Anonymous
Solve the equation to find the value of x : log10 (3x+2) – 2 log10 x = 1– log10 (5x – 3)
Answers
Answered by
oobleck
assuming base 10 for the logs,
log(3x+2) – 2 log(x) = 1– log(5x – 3)
log(3x+2) – 2 log(x) + log(5x – 3) = 1
Now, using the basic properties of logs, we have
log (3x+2)(5x-3)/x^2 = 1
(3x+2)(5x-3) = 10x^2
(6x+5)(x-1) = 0
x = - 5/6 or 1
But logx is undefined for x<0, so x=1 is the only solution.
check:
log5 - 2log1 = 1-log2
log5+log2 = 1
log10 = 1
true!
log(3x+2) – 2 log(x) = 1– log(5x – 3)
log(3x+2) – 2 log(x) + log(5x – 3) = 1
Now, using the basic properties of logs, we have
log (3x+2)(5x-3)/x^2 = 1
(3x+2)(5x-3) = 10x^2
(6x+5)(x-1) = 0
x = - 5/6 or 1
But logx is undefined for x<0, so x=1 is the only solution.
check:
log5 - 2log1 = 1-log2
log5+log2 = 1
log10 = 1
true!
Answered by
Sophia Samuel
More questions pls..
Answered by
ADA
log10 (2x + 5x _ 2)= 1
Answered by
ADA
log10 (2x + 5x _ 2) = 1
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