document.write( "Question 75928: I am trying to solve the following equation:\r
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document.write( "log_5 x/125 = log_7(49x^3)\r
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document.write( "This is what I have tried...where am I going wrong?\r
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document.write( "log_5 (x) - log_5 (5^3) = log_7(7^2) + log_7 (x^3)
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document.write( "log_5 (x) - 3 = 2 + 3 log_7 (x)
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document.write( "-5 = x(3 log_7 - log_5)
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document.write( "Please help....when I now divide...I get x to be undefined?
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document.write( "Thank you \n" );
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Algebra.Com's Answer #54521 by scott8148(6628) You can put this solution on YOUR website! You're fine until the last step...you \"factor out\" the argument of the two logs, leaving undefined values (log_7 and log_5 of what?)...You need to get the bases of the logs the same (like common denominators for fractions).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-5=3log_7(x)-log_5(x)... this equals \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "multiplying by log 7 and log 5 and factoring gives...-5 log7 log5 = logx (3 log5 - log7)...dividing by (3 log5 - log7) gives |