How do you Solve for x: # log_5(4x-3)=2#?

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Answer 1 · 2018-04-25T19:13:18

#x=7#

Remember: #log_b a=c# means #b^c=a#

So #log_5(4x-3)=2#
means 5^2=4x-3#

#rArr 25=4x-3#

#rArr 28=4x#

#rArr 7=x#

Answer 2 · 2018-04-25T19:14:00

#x=7#

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

#•color(white)(x)log_b x=nhArrx=b^n#

#log_5(4x-3)=2#

#rArr4x-3=5^2=25#

#rArr4x=25+3=28rArrx=28/4=7#

Answer 3 · 2018-04-25T19:18:36

# x=7#.

Recall the definition of #log# function given below :

#log_b a=m hArr b^m=a#.

#:. log_5 (4x-3)=2#,

#rArr 5^2=(4x-3)#.

# rArr 25=4x-3#.

# rArr 25+3=4x#.

# rArr 28=4x#.

# rArr x=28/4#.

# rArr x=7#.

This root also satisfy the given eqn.

#:. x=7# is the desired solution.