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log(7)3=K 7^(3y-1)=21 howw
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\[\log _{7}3=k\] \[7^{3y-1}=21\]
What's the question? Get k and y?
y in terms of K
sorry
\[7^{3y-1}=21\] \[7^{3y-1}=7*3\] \[7^{3y-2}=3\] \[log{7^{3y-2}}=log{3}\] \[(3y-2)log{7}=log{3}\] \[(3y-2)=log{3}/log(7)\] \[3y-2=log_7{3}\] \[3y-2=k\] \[3y=k+2\] \[3y=(k+2)/3\]
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Last one is \[y=(k+2)/3\]
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