How do you solve #(1/7)^(y-3)=343#?

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Answer 1 · 2016-11-08T18:15:57

Please see the explanation for steps leading to the answer: #y =0#

Given:

#(1/7)^(y - 3) = 343#

Take the natural logarithm on both sides:

#ln((1/7)^(y - 3)) = ln(343)#

Use the property of logarithms #log_b(a^c) = (c)log_b(a)# on the left side:

#(y - 3)ln(1/7) = ln(343)#

Divide both sides by #ln(1/7)#:

#y - 3 = ln(343)/ln(1/7)#

Use the property of logarithms #log_b(1/a) = -log_b(a)# on the #ln(1/7)#:

#y - 3 = ln(343)/-ln(7)#

Add 3 to both sides:

#y = 3 - ln(343)/ln(7)#

#y = 0#

Answer 2 · 2016-11-08T19:42:55

#y=0#

#(1/7)^(y-3)=343#

now # 7^3=343#

& # (1/7)^-1=7#

#:.(1/7)^(y-3)=7^(3-y)#

#7^(y-3)=7^3=343#

#=>y-3=0#

#:.y=0#