How do you solve #10^(4x-1) = 5000#?

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Answer 1 · 2018-03-02T19:00:45

#x=ln(5000)/(1ln(4))+1/4#

Apply the natural logarithm to both sides:

#ln(10^(4x-1))=ln(5000)#

#ln(10^(4x-1))=(4x-1)ln(10),# as the exponent property for logarithms tells us that #ln(x^y)=yln(x)#.

#ln(10^(4x-1))=ln(5000)hArr(4x-1)ln(10)=ln(5000)#

Solve for #x:#

#((4x-1)cancelln(10))/cancelln(10)=ln(5000)/ln(10)#

#4xcancel(-1+1)=ln(5000)/ln(10)+1#

#(cancel(4)x)/cancel4=ln(5000)/(4ln10)+1/4#