Can someone pls help me
write 3lgx + 2-lgy as a single logarithm

log(x^3*y^(-2))

Step-by-step explanation:

two log rules you need to know for this question:

alogb = log(a^b)

logx + logy = logxy

you can start by getting rid of the term in front of the log, or a

also is it 2 minus or 2 times negative? im assuming it's 2 times negative because it is more likely

log (x^3) + log (y^-2) = log (x^3*y^-2)

option a is correct.

step-by-step explanation:

scale factor: it is used to find proportional measurements.

proportions means having the same ratio.

scale factor is the ratio of the model measurement to the actual measurement in a simplest form.

or we can say that the ratio of any two corresponding lengths in two similar geometric figures

if the scale factor is more than 1, then it is an enlargement,

if the scale factor less than 1, then it is a reduction.

given the two geometric figure:

by definition of proportions, we have

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w