Which of the following logarithmic equations has no solution? log4(x + 2) = log4(6 - x) log4(-x) = 1 log4(x + 2) = 1 log4(x+2) = log4(x+6)

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Which of the following logarithmic equations has no solution? log4(x + 2) = log4(6 - x) log4(-x) = 1 log4(x + 2) = 1 log4(x+2) = log4(x+6)

Mathematics
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try to solve for x hint:hint: \[\huge\rm log_b a=\log_b c\] \[\large\rm \cancel{log_b} a=\cancel{\log_b }c\] a=c
bases of the log are same both sides so you cancel out log_B
oh so it would just be x+2=6-x

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also try finding the log of -4 on your calculator
- or the log of any negative for that matter
for 2nd) 3rd) you need to convert log to exponential form |dw:1439137679880:dw| after that you can plugin the x value to check ur answer
non real answer, so that means it would be log4(-x) = 1 , right?
so like a first step i think you need to know where is deffined the logaritmic function ? ok ?
oh second thoughts x could be negative - in which case -x is positive - so its not the second option
\(\color{blue}{\text{Originally Posted by}}\) @arnavmishra oh so it would just be x+2=6-x \(\color{blue}{\text{End of Quote}}\) yes right but you can solve for x there is a solution.
i ended up with log4(x + 2) = 1, is that the right answer?
well solve that for x x= ??
for example first one \[x+2=6-x\] solve for x move the -x to the left side and 2 to the right side \[x+x=6-2\]\[2x=4\] x=2 now plugin the 2 into the original equation for x \[\log_4(2+2)=\log_4(6-2)\] both sides are equal so 2 is a solution
  • phi
the last choice \[ \log_4(x+2) = \log_4(x+6) \] make each side the exponent of the base 4: \[ 4^{\log_4(x+2) }= 4^{\log_4(x+6)}\\ x+2= x+6 \] (we could have "jumped" to this equation) now add -x to both sides: \[x-x+2 = x-x + 6\\ 2= 6\] we know 2 is not 6, so something is wrong... namely there is no solution to the original expression.

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