Asked by jay
                use the properties of logarithms for the expression.
5^log5(7x)
            
        5^log5(7x)
Answers
                    Answered by
            bobpursley
            
    Is there an equal sign missing?  Idont know what you want to do here.
a^loga (yyy)=yyyy
    
a^loga (yyy)=yyyy
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