3logy^3 + log y^2 I had this on a test I put logy^18 was the answer
3 logy^6 ?
3Log y^3 + log y^2
log (y^9*y^2)
log y^11
check: let y= 11
3log 10^3 + log 10^2= 9+2=11
oh you add the exponents I multipied, Thanks
To simplify the expression 3logy^3 + logy^2, we can use the logarithmic property that states loga(x) + loga(y) = loga(xy).
In this case, we have 3logy^3 + logy^2. Since both terms have the same base (log), we can add the coefficients and multiply the bases together.
First, let's simplify the coefficients:
3log y^3 can be written as log y^3 * y^3 = log y^9
log y^2 remains unchanged.
Now we can rewrite the expression as log y^9 + log y^2.
Using the logarithmic property mentioned earlier, we get:
log y^9 * y^2 = log y^18.
So, your answer of logy^18 is indeed correct. Well done!