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

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Answer 1 · 2015-07-08T14:10:46

use taking logarithm on both sides

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

moving (-3) to RHS

#(1/7)^x=343+3#

now applying ln on both sides

#ln((1/7)^x)=ln(346)# (eq 1)

now recall,

#lna^b=blna#

applying this to eq 1

#x.(ln(1/7))=ln(346)#

now simplifying the eq we get

#x=ln(346)/(ln(1/7))#

for solving denominator recall

#ln(x / y) = ln(x) - ln(y)#

we get

#x=ln(346)/(ln1-ln7)#

#ln1=0#

now calculating ln346 and ln7 we get

#x=5.8377304472/(-1.9459101491)#

#x=-2.99999999995#

upon rounding it up we get

#x=-3#

any correction is welcome

Cheerio!