keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
3x^2+11x-4=10x-1
3x^2+11x-10x-4+1=0
3x^2+x-3=0
Δ=1^1-4*3*(-3)=1+36=37
x1=(-1+V37)/6
x2=( -1-V37)/6
Answer: 52.77876 in
Explanation: C=2πr=2·π·8.4≈52.77876 in
We know the equation of a line is y=mx+b. We know the slope (m) is -2 and all we need is the intercept (b).
Since we know one point, we can plug in 6 for x and 3 for y (along with -2 for m) in our equation and then solve for b.
3= -2*6 + b
3= -12 + b
15 = b
Since we now know the intercept we can now write our equation:
y= -2x + 15