Asked by james
Solve for x in the following equations. (4 marks)
log3(x - 5) + log3(x + 3) = 2 <--- both log3.. 3 is subscript
log7x + log7(x - 1) = log72x <--- all log7.. 7 is subscript.
log3(x - 5) + log3(x + 3) = 2 <--- both log3.. 3 is subscript
log7x + log7(x - 1) = log72x <--- all log7.. 7 is subscript.
Answers
Answered by
Reiny
use your log rules:
log (AB) = logA + logB
and
log (A/B) = logA - logB
1st one:
log<sub>3</sub> [(x-5)(x+3)] = 2
by definition of logs
(x-5)(x+3) = 3^2 = 9
x^2 - 2x -24=0
(x-6)(x+4) = 0
x = 6 or x=-4 , but x=-4 would make log<sub>3</sub>(x+3) undefined, so
x = 6
2nd:
log<sub>7</sub>(x) + log<sub>7</sub>(x-1) - log<sub>7</sub>(2x) = 0
log<sub>7</sub>[x(x-1)/(2x)] = 0
log<sub>7</sub>[(x-1)/2] = 0
(x-1)/2 = 7^0
(x-1)/2 = 1
x-1=2
x = 3
log (AB) = logA + logB
and
log (A/B) = logA - logB
1st one:
log<sub>3</sub> [(x-5)(x+3)] = 2
by definition of logs
(x-5)(x+3) = 3^2 = 9
x^2 - 2x -24=0
(x-6)(x+4) = 0
x = 6 or x=-4 , but x=-4 would make log<sub>3</sub>(x+3) undefined, so
x = 6
2nd:
log<sub>7</sub>(x) + log<sub>7</sub>(x-1) - log<sub>7</sub>(2x) = 0
log<sub>7</sub>[x(x-1)/(2x)] = 0
log<sub>7</sub>[(x-1)/2] = 0
(x-1)/2 = 7^0
(x-1)/2 = 1
x-1=2
x = 3
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