Asked by jessica

log 1/x + 3log2=log32
Answered by bobpursley
adding logs is same as multplying

log((2^3)/x)=log32

take the antilog of each side.

8/x=32
solve for x
Answered by ibrahim kabir
log1/x+3log2=log32 log1/x+log2^3=log32 log1/x+log8=log32 according to the law of logarithm +will change to * therefore log(1/x*8)=log32 since the base is equal the they will cancel each other in the equation (1/x*8)=32 (8/x)=32 clear bracket and cross multiply 8=32x divide both sides by 32 to make it x=1/4
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