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6
4 (a) Given 
log_(2)3=p and 
log_(2)5=q, express and simplify 
log_(8)((2)/(5))+log_(3)2 in terms of 
p and 
q.
[3]

66\newline44 (a) Given log23=p \log _{2} 3=p and log25=q \log _{2} 5=q , express and simplify log825+log32 \log _{8} \frac{2}{5}+\log _{3} 2 in terms of p p and q q .\newline[33]

Full solution

Q. 66\newline44 (a) Given log23=p \log _{2} 3=p and log25=q \log _{2} 5=q , express and simplify log825+log32 \log _{8} \frac{2}{5}+\log _{3} 2 in terms of p p and q q .\newline[33]
  1. Change Base to Base 22: Change the base of log8\log_{8} to base 22 using the change of base formula: logba=logkalogkb\log_{b} a = \frac{\log_{k} a}{\log_{k} b}.
    log8(25)=log2(25)log28.\log_{8} \left(\frac{2}{5}\right) = \frac{\log_{2} \left(\frac{2}{5}\right)}{\log_{2} 8}.
  2. Simplify Log Base 22 of 88: Simplify log28\log_{2} 8 since 88 is 22 cubed, log28\log_{2} 8 equals 33.\newlinelog8(25)=log2(25)3.\log_{8} \left(\frac{2}{5}\right) = \frac{\log_{2} \left(\frac{2}{5}\right)}{3}.
  3. Split Log Base 22 of (25)(\frac{2}{5}): Split log base 22 of (25)(\frac{2}{5}) into log base 22 of 22 minus log base 22 of 55 using the quotient rule.\newlinelog8(25)=(log22log25)/3\log_{8}(\frac{2}{5}) = (\log_{2} 2 - \log_{2} 5) / 3.
  4. Substitute Given Values: Substitute the given values: log23=p\log_{2} 3 = p and log25=q\log_{2} 5 = q.\newlinelog8(25)=1q3\log_{8} \left(\frac{2}{5}\right) = \frac{1 - q}{3} because log22=1\log_{2} 2 = 1.
  5. Change Base to Base 22: Change the base of log32\log_{3} 2 to base 22 using the change of base formula.log32=log22log23\log_{3} 2 = \frac{\log_{2} 2}{\log_{2} 3}.
  6. Substitute Given Value: Substitute the given value for log23\log_{2} 3.\newlinelog32=1p\log_{3} 2 = \frac{1}{p} because log22=1\log_{2} 2 = 1.
  7. Combine Expressions: Combine the two expressions.\newlinelog8(25)+log32=1q3+1p\log_{8}\left(\frac{2}{5}\right) + \log_{3}2 = \frac{1 - q}{3} + \frac{1}{p}.
  8. Simplify Expression: Simplify the expression.\newlineFinal expression in terms of pp and qq: 1q3+1p\frac{1 - q}{3} + \frac{1}{p}.

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