# math

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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.

• math -

log (AB) = logA + logB
and
log (A/B) = logA - logB

1st one:
log3 [(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 log3(x+3) undefined, so

x = 6

2nd:
log7(x) + log7(x-1) - log7(2x) = 0

log7[x(x-1)/(2x)] = 0
log7[(x-1)/2] = 0
(x-1)/2 = 7^0
(x-1)/2 = 1
x-1=2
x = 3

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