Hi. Hoping you doing great this evening. I know you must bepreparing for bed. Pl

Question

Hi. Hoping you doing great this evening. I know you must bepreparing for bed. Please can you help me to solve thisproblem? 1a) I want to solve this equation by expressing each side asapower of the same base and then equating exponents:-2(1+2x) = 32.    (b) I want to solve thelogarithmic equation. I was asked to make sure to reject anyvalue that is not in the domain of the original logarithmicexpressions. I need to give the exact answer:- log 3x = log 5 +log(x-4). Please help they are due tonight. Thanks. Hi. Hoping you doing great this evening. I know you must bepreparing for bed. Please can you help me to solve thisproblem? 1a) I want to solve this equation by expressing each side asapower of the same base and then equating exponents:-2(1+2x) = 32.    (b) I want to solve thelogarithmic equation. I was asked to make sure to reject anyvalue that is not in the domain of the original logarithmicexpressions. I need to give the exact answer:- log 3x = log 5 +log(x-4). Please help they are due tonight. Thanks.

Explanation / Answer

1a.. 2^(1+2X)=32 2^(1+2X)=2^5.....as 2^5=32 1+2x=5..... 2x=5-1 x=4/2=2 b..log 3x=log 5 +log (x-4) log 3x=log (5.(x-4)) log 3x=log (5x-20) log 3x - log (5x-20)=0 log(3x/(5x-20))=0 (3x/5x-20)=A^0 where A is a constant (3x/5x-20)=1 3x=(5x-20).1 3x=5x-20 -2x= -20 x= (-20/-2) = 10 (ans)

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