# Answer to Question #2840 in Algebra for Keyla

solving exponential equations with logarithms...
3^x=17
12^x=17
9^x=4.9
6^x= 0.51
18^2x=26
2^5x=34

x = log_{3} 17

x = log_{12} 17

x = log_{9} 4.9

x = log_{6} 0.51

x = ½ log_{18} 26

x = 1/5 log_{2} 34

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