Solve: 10^logx + log4 ln(2x^2 + 4x) = 5 + ln2x

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neela | High School Teacher | (Level 3) Valedictorian

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To solve 

(i)  10^logx +log4 

(ii) ln(2x^2+4x)=5+ln2x.

(i) 10^logx +log4 is not an expression (not an equation). It could be simplified only.

First term: Let 10^logx = y. So by definition logx = log y. Or y = x. Therefore 10^logx = x.

So 10 ^logx +log4 = x+log4 .If this expression is equal to a number n,

x+log 4 = n. So

x = n- log4.

ln(2x^2+4x) = 5+ln2x.

ln(2x^2+4x) -ln2x = 5

ln{(2x^2+4x)/2x} = 5,as lna-lnb = ln(a/b)

ln(x+2) = 5 = lne^5.

x+2 = e^5, as

x = (e^5)-2

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