Common Log Notes-completed

Math III Name: Notes: Common Log/Change of Base Formula Base 10 logarithms are called common logarithms. Common logs are often written without the subscript 10. [log_{10}x\rightarrow log~x] The log key on the calculator only evaluates base 10 logs. I. Evaluate. (Round to 4 decimal places) (log~110) 2) (log~7.5) 2.0414 *Do all steps in calc. Round at the end What if the base is not 10? Take the log of both sides. 0.8751 II. Solve. (Round to 4 decimal places) (log_{10}x=0.33) 4) (log_{10}x=2.5) 5) (10^{x}=4) \[10^{0.33}=x\] \[10^{2.5}=x\] \[log_{10}4=x\] \[x\approx2.1380\] \[x=0.6021\] \[x\approx316.2278\] ex for #5 \[og~10^{x}=log~4\] \[log~10=log~4\] \[x=\frac{log~4}{log~10}\] *! (5^{2z}=19) In Calc 7) (5^{s-2}=3) 8) (2^{\overline{4}}=15) [\frac{l~cog~19}{log~5}\rightarrow Ans] [log~5^{x-2}=log3] Ans/2 [(x-2)log~5=log_{2}g^{3}] Final [x-2=\frac{log^{3}}{log~6}] Ans [X=\frac{log^{3}}{log5}+2] [x\approx2.6826] \[log~5^{2x}=log~19\] \[2~log~5=log~19\] \[2x=\frac{log14}{log5}\] \[2x=1.82....\] \[x\approx0.9/47\] \[log~2^{x/4}=log_{x=}^{15}\] \[\frac{x}{4}log~2=log^{15}\] \[\frac{x}{4}=\frac{log~15}{log~2}\] \[x=4\cdot\frac{log~15}{log^{2}}\] \[x\approx15.6276\] (4^{2x+3}+5=31) *10) (3^{t+5}=2^{4-x}) [+2x+3=26] [log~3^{x+5}=log~2] [(x+5)log^{3}=(4-x)log^{2}] [x~log~3+5~log~3=4~log^{2}-x~log^{2}] [x~log~3+x~log~2=4~log^{2}-5~log^{3}] [x(log~3+log~2)=4~log^{2}-5~log^{3}] -x \[2x\approx-0.6498\] \[x\approx-0.3249\] \[log~4^{2x+3}=log~26\] \[x=\frac{4log^{2}-5~log^{3}}{log~3+log^{2}}\] \[(2x+3)log~4=log~26\] \[X\approx-1.5/83\] \[2x+3=\frac{log~26}{log~4}\] \[2x=\frac{log~26}{log~4}-3\] \[3~log~x=12\] 12) \(log~2x-log~4=3\) \[log_{10}x=4\] \[log\frac{2x}{4}=3\] \[10^{4}=x\] \[10^{3}=\frac{x}{2}\] 10000=X \[1000=\frac{x}{2}\] \[2000=x\] ← Circle method #3-5 Condensell circle Math III (log~x+log~7=2) 14) (8^{x-1}=13) \[log~7x=2\] \[log~8^{x-1}=log~13\] \[(x-1)log~8=log^{13}\] \[x-1=\frac{log^{13}}{log\theta}\] \[x=\frac{log13}{log\theta}+1\] \[x\approx2.2335\] \[10^{2}=7x\] \[100=7x\] \[x=\frac{100}{7}\] (3log(x+1)=6) \[log(x+1)=2\] \[10^{2}=x+1\] \[x=99\] \[100=x+1\] Change of Base Formula The Change of Base Formula allows you to write equivalent log expressions that have different bases. For all positive numbers, a and n. where (a\ne1) [log_{a}n=\frac{log~n}{log~a}]   Express each log in terms of common logs. Then approximate the value to 4 decimal places. log, 12 [\frac{log~12}{log~5}\approx1.5440] (log_{4}25) [\frac{log~25}{log^{4}}\approx2.3219] log, 5 [\frac{log~5}{log~7}\approx0.82~7|] (log_{3}42) [\frac{log^{42}}{log^{3}}\approx3.4022]