Question 829853
find x
1) log (x-1) to base 2 + log (x+1) to base 2 = 3
{{{log(2,(x-1)) + log(2,x+1)) = 3 = log(2,8)}}}
{{{log(2,(x-1)*(x+1)) = log(2,8)}}}
{{{(x-1)*(x+1) = 8}}}
{{{x^2 - 1 = 8}}}
{{{x^2 = 9}}}
x = 3
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x = -3 is rejected, gives log(-4)
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2) log (4-2x)=log(x+1)
Simple
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3) log (9-2x) to base 1/3=log1/3(x-6)
Is 1/3 the base of both?
Use log(1/3,(9-2x)) with 3 sets of braces { } to render legibly.
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