Adding Logs with the Same Base
Read this topicExplanations 2
- Okay, so, we're given two numbers: 1 and 2. Awesome! Now let's write the numbers 1 and 2 in log base 10. Remember, log base 10 is the standard base, so we can just write log(x)…
By Freya Edholm, Tristan Pollner, Tom Shen, Expii Team
- Remember that logarithms are just exponents. For example, we know that log28=3\log_{2}{8}=3log28=3, where 3 is the exponent. This means that when we add logarithms, we are ess…
By Neha Prasad, Alex Federspiel
Problems 21
- Compute 1log2100!+⋯+1log100100!\dfrac{1}{\log_2{100!}} + \cdots + \dfrac{1}{\log_{100}{100!}}log2100!1+⋯+log100100!1.
- What is 2⋅log4(8)2 \cdot \log_4(8)2⋅log4(8)?
- What is log6(3)+log6(12)\log_6(3) + \log_6(12)log6(3)+log6(12)?
- Fix b,x,y,zb, x, y, zb,x,y,z such that logb(x)+logb(y)=z\log_b(x) + \log_b(y) = zlogb(x)+logb(y)=z. In terms of bbb and zzz, what is xyxyxy equal to?
- If log6(x)+log6(4)=2\log_6(x) + \log_6(4) = 2log6(x)+log6(4)=2, find xxx.
- Solve for xxx if log7(x−3)+log7(x+2)=log714\log_{7}(x-3) + \log_{7}(x+2) = \log_{7}14log7(x−3)+log7(x+2)=log714.
- Simplify 3log2(x)+log2(x+3)+log2(y)3\log_{2}(x)+\log_{2}(x+3)+\log_{2}(y)3log2(x)+log2(x+3)+log2(y).
- Let logb(2)=0.3869,logb(3)=0.6131, and logb(5)=0.8982\log_{b}(2) = 0.3869, \log_{b}(3)= 0.6131, \text{ and } \log_{b}(5) = 0.8982logb(2)=0.3869,logb(3)=0.6131, and logb(5)…
- Find the value of log10(20)+log10(50)\log_{10}(20)+\log_{10}(50)log10(20)+log10(50).
- Solve the logarithmic equation log2(x+2)+log2(3)=log2(27)\log_{2} (x+2) + \log_{2} (3) = \log_{2}(27)log2(x+2)+log2(3)=log2(27) for xxx.
- Solve the logarithmic equation: log3(2x−7)+log32=2\log _{3} (2x-7)+\log_{3}2= 2log3(2x−7)+log32=2
- Solve for xxx: log3(1+log3(2x−7))=1\log_{3}(1+\log_{3}(2^{x}-7)) = 1log3(1+log3(2x−7))=1
- log5+log40+log20=?\log 5 +\log 40 + \log 20 = ?log5+log40+log20=?
- Solve the logarithmic equation: log2(x+2)+log23=log227\log_{2}(x+2)+ \log_{2}3=\log_{2}27log2(x+2)+log23=log227
- Solve for xxx: log3x+log3(x−2)=log3(x+10)\log_{3}x+ \log_{3}(x-2)= \log_{3}(x+10)log3x+log3(x−2)=log3(x+10)
- Solve for xxx: log2(3x+1)+log2(−x−5)=log28\log_{2} (3x+1) + \log_{2} (-x-5) = \log_{2}8log2(3x+1)+log2(−x−5)=log28
- Solve for xxx: log(5x+4)+log7=2\log (5x+4) + \log 7 = 2log(5x+4)+log7=2
- Simplify: 4log7x+log7(y3+2)+log7(y+4)4\log_{7} x +\log_{7} (y^{3}+2)+\log_{7}(y+4)4log7x+log7(y3+2)+log7(y+4)
- Find the value of log749+log77\log_{7}49 + \log_{7}7log749+log77.
- Let loga(9)=0.7891,loga(2)=0.0231, and loga(4)=0.4556\log_{a}(9)= 0.7891, \log_{a}(2)= 0.0231, \text{ and } \log_{a}(4)=0.4556loga(9)=0.7891,loga(2)=0.0231, and loga(4)=0.…
- Solve for xxx: log5(x+1)+log5(2x+3)=log513\log_{5}(x+1)+\log_{5}(2x+3) = \log_{5}13log5(x+1)+log5(2x+3)=log513