SOLUTION: Please, how do I solve this equations: log49^343 and log10^(2x+1) - log10^(3x-2)=1. The steps will help a lot. Thanks.

Question 512199: Please, how do I solve this equations: log49^343 and log10^(2x+1) - log10^(3x-2)=1. The steps will help a lot.
Thanks.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
log49^343 and log10^(2x+1) - log10^(3x-2)=1
**
log49^343 + log10^(2x+1) - log10^(3x-2)=1
343log49+(2x+1)log10-(3x-2)log10=1
log of base=1, so log 10=1
343log49+(2x+1)-(3x-2)=1
2x+1-3x+2=1-343log49
-x=-2-343log49
x=2+343log49=2+579.7373=581.7373
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