Understand logarithms as inverses of exponentials
A logarithm answers: "What exponent gives me this result?"
logb(x)=y ⟺ by=x\log_b(x) = y \iff b^y = xlogb(x)=y⟺by=x
log2(8)=3\log_2(8) = 3log2(8)=3 means "2 to what power equals 8?" Answer: 3
log(x)=log10(x)\log(x) = \log_{10}(x)log(x)=log10(x)
ln(x)=loge(x)\ln(x) = \log_e(x)ln(x)=loge(x)
logb(1)=0\log_b(1) = 0logb(1)=0 (because b0=1b^0 = 1b0=1) logb(b)=1\log_b(b) = 1logb(b)=1 (because b1=bb^1 = bb1=b) logb(bx)=x\log_b(b^x) = xlogb(bx)=x blogb(x)=xb^{\log_b(x)} = xblogb(x)=x
logb(xy)=logb(x)+logb(y)\log_b(xy) = \log_b(x) + \log_b(y)logb(xy)=logb(x)+logb(y)
logb(xy)=logb(x)−logb(y)\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)logb(yx)=logb(x)−logb(y)
logb(xn)=nlogb(x)\log_b(x^n) = n\log_b(x)logb(xn)=nlogb(x)
logb(x)=loga(x)loga(b)=ln(x)ln(b)\log_b(x) = \frac{\log_a(x)}{\log_a(b)} = \frac{\ln(x)}{\ln(b)}logb(x)=loga(b)loga(x)=ln(b)ln(x)
To solve 2x=102^x = 102x=10:
y=logb(x)y = \log_b(x)y=logb(x) is the reflection of y=bxy = b^xy=bx over y=xy = xy=x