I'm all about that base, 'bout that base, no treble
( lo g x 2 ) 2 + lo g 1 5 = 1 + lo g 3 − lo g 2 , x = ?
Solve for positive real value of x .
Details and Assumptions :
- Although the actual value is not specified, the base of all of the logarithms given above are the same.
The answer is 1.
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2 solutions
( l o g x 2 ) 2 + l o g 1 5 = 1 + l o g 3 − l o g 2
⇒ ( l o g x 2 ) 2 + l o g ( 3 ∗ 5 ) = 1 + l o g 3 − l o g 2
⇒ ( l o g x 2 ) 2 + l o g ( 3 ) + l o g 2 1 0 = 1 + l o g 3 − l o g 2
⇒ ( l o g x 2 ) 2 + l o g 3 + l o g 1 0 − l o g 2 = 1 + l o g 3 − l o g 2
⇒ ( l o g x 2 ) 2 + l o g 3 + 1 − l o g 2 = 1 + l o g 3 − l o g 2
⇒ ( l o g x 2 ) 2 = 0
x = 1
x can also be − 1 .
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But -1 is not the domain of Logarithm, I am right?
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If x = − 1 , then lo g x 2 = lo g ( − 1 ) 2 = lo g 1 = 0 . (Also, note that the LaTeX code for logarithm is \log.)
That is correct
Note that the problem says "positive real x". The more unusual part to me is that if you do not assume base 10, you get a complicated expression with domain (0,1) U [10,infinity)
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At the time I posted the comment above, the problem did not include the condition that x had to be positive. The problem has been edited since then.
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@Jon Haussmann – Oh I didn't realize that. Good job by you looking at how the problem is actually written instead of just going by "what usually happens"
A much easier solution would be this: Using the logarithm laws: LogMN = Log M + LogN and LogM/N = LogM - LogN, The equation can be simplified to log(15x^4) = 1 + log(3/2). Since 1 can be rewritten as log10, log(15x^4) = log10 + log(3/2) -> log(15x^4) = log(30/2) -> log15x^4 = log15, so 15x^4=15 -> x^4=1 -> x=1
(log x^2 )^2 + log 15 = 1 + log 3 - log 2
log x^4 + log 15 = log (10 * 3) / 2
log x^4 + log 15 = log 15
log x^4 = 0
x^4 = 10^0
x = 1