Asked by Anonymous
Express as a single logarithm.
1/2 log base b of (x-2)- log base b of y + (3 log base b of z) = log base 6 of (x-2)^1/2
1/2 log base b of (x-2)- log base b of y + (3 log base b of z) = log base 6 of (x-2)^1/2
Answers
Answered by
Reiny
(1/2) log<sub>b</sub>(x-2) - log<sub>b</sub>y + 3log<sub>b</sub>z = log<sub>6</sub>(x-2)^(1/2)
log<sub>b</sub> [ (x-2)^(1/2) z/y] = log<sub>6</sub>(x-2)^(1/2)
log<sub>b</sub> ( (z/y)√(x-2) ) = log<sub>6</sub> √(x-2)
are you sure the right side was not supposed to be
log<sub>b</sub> ?
log<sub>b</sub> [ (x-2)^(1/2) z/y] = log<sub>6</sub>(x-2)^(1/2)
log<sub>b</sub> ( (z/y)√(x-2) ) = log<sub>6</sub> √(x-2)
are you sure the right side was not supposed to be
log<sub>b</sub> ?
Answered by
Reiny
oops, didn't notice the 3 coefficient until I checked the display of this post.
should have been: z^3 in the last two lines.
log<sub>b</sub> (x^3/y)log<sub>b</sub>(x-2) = log<sub>6</sub> √(x-2)
should have been: z^3 in the last two lines.
log<sub>b</sub> (x^3/y)log<sub>b</sub>(x-2) = log<sub>6</sub> √(x-2)
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