anonymous
  • anonymous
log8(x+1)=2/3 (a small 8)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\log_ab = c \iff a^c =b \]
anonymous
  • anonymous
\[x+1=8^{\frac{2}{3}}\] \[x+1=\sqrt[3]8^2\] \[x+1=2^2\] \[x+1=4\] \[x=3\]
anonymous
  • anonymous
\[\log_8(x+1)=\frac{2}{3}\] now we make both of the sides an exponent of 8. It'll look like this \[8^{\log_8(x+1)}=8^{\frac{2}{3}}\] the left side simply becomes x+1 because of a common cancellation rule of logs. In general, \[a^{\log_ab}=b\] since log base a of b is the exponent we must raise a to in order to get b, if we make log base a of b an exponent of a, we get b the left side is \[8^\frac{2}{3}=\sqrt[3]{8}^2=2^2=4\] therefore, x+1=4, and x=3

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