How do you evaluate the expression $6^{{log}_{6} (3 x + 2)}$ $6^(log_6(3x+2))$ ?

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Answer 1 · 2016-10-29T10:07:28

$6^{{log}_{6} (3 x + 2)} = 3 x + 2$ $6^(log_6(3x+2))=3x+2$

Let $6^{{log}_{6} (3 x + 2)} = u$ $6^(log_6(3x+2))=u$

then using definition of logarithm that ${log}_{a} m = b \Rightarrow m = a^{b}$ $log_a m=b rArr m=a^b$ , we have

${log}_{6} u = {log}_{6} (3 x + 2)$ $log_6 u=log_6(3x+2)$

i.e. $u = 3 x + 2$ $u=3x+2$

Hence $6^{{log}_{6} (3 x + 2)} = 3 x + 2$ $6^(log_6(3x+2))=3x+2$

How do you evaluate the expression 6^(log_6(3x+2))6log6(3x+2)6^(log_6(3x+2))?