How do you solve #log_5x=3#?

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Answer 1 · 2018-08-14T18:44:06

#log_5 x=3 <=>x=5^3=125#

Here ,

#log_5 x=3#

We know that , for #X , YinRR^+ anda inRR^+ -{1}#

#color(red)(log_aX=Y <=>X=a^Y to" Definition"#

#:.log_5 x=3 <=>x=5^3=125#

Answer 2 · 2018-08-14T20:03:19

#x=125#

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)log_b x=nhArrx=b^n#

#log_5 x=3rArrx=5^3=125#