Question 177517
Hi, Hope I can help
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{{{ log (2,(x+20))- log (2,(x+2)) = log (2,x) }}}
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You are correct
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If you subtract the same number log by the log, it is the same as one complete log as a fraction/division
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{{{ log (2,((x+20)/(x+2))) = log (2,x) }}}
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This is where you got stuck, from here we can use a really easy method called "undo"
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{{{ log (2,((x+20)/(x+2))) = log (2,x) }}}
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Since this equation has the same base/log (2) on both sides, we can just "undo" , get rid of, erase, cross out the logs
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 {{{ log (2,((x+20)/(x+2))) = log (2,x) }}} = {{{ (x+20)/(x+2) = x }}}
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Now just solve for "x"
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{{{ (x+20)/(x+2) = x }}}, we will multiply each side by (x+2) to get rid of the fraction
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{{{ (x+20)/(x+2) = x }}} = {{{ (x+2)((x+20)/(x+2)) = x(x+2) }}} = {{{ highlight(x+2)((x+20)/highlight(x+2)) = x(x+2) }}} = {{{ cross(x+2)((x+20)/cross(x+2)) = x(x+2) }}} = {{{ x+20 = x(x+2) }}}, we will use distribution on the right side
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{{{ x+20 = x(x+2) }}} = {{{ x+20 = highlight(x)(highlight(x)+2) }}} = {{{ x+20 = highlight(x)(x+highlight(2)) }}} = {{{ x + 20 = x^2 + 2x }}}
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We will now move "x + 20" to the right side
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{{{ x + 20 = x^2 + 2x }}} = {{{ x - x + 20 - 20= x^2 + 2x - x - 20 }}} = {{{ 0 = x^2 + x - 20 }}}
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{{{ 0 = x^2 + x - 20 }}} or {{{ x^2 + x - 20  = 0}}}, this is a quadratic equation, we can solve this equation by factoring
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{{{ x^2 + x - 20  = 0}}}
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To find the factors, we have to have multiples of "-20", and the multiples have to add up to the middle term, or "1"
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Factors of (-20)
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(2,(-10)) No
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(4,(-5)) No
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(5,(-4)) Yes, {{{ 5 + (-4) = 1 }}} (these factors add up to "1")
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Since there was only {{{ x^2 }}}, that means there are only "x's"
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(x)(x), you would then put the factors inside
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(x+5)(x-4) are the factors, you can check using the foil method, but I already checked
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{{{ (x+5)(x-4) = 0 }}}, now we can solve for "x", you can find "x" by putting each factor equal to zero
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{{{ x + 5 = 0 }}}, now move "5" to the right side
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{{{ x + 5 - 5 = 0 - 5 }}} = {{{ x = (-5) }}}
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{{{ x - 4 = 0 }}}, now move (-4) to the right side
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{{{ x - 4 + 4 = 0 + 4 }}} = {{{ x = 4 }}}
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{{{ x = (-5) }}}
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{{{ x = 4 }}}
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These are your answers, but, you have to make sure and plug your answers into the original equation, and make sure that the numbers don't make a negative, because you can't take the log of a negative number
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(-5), {{{ log (2,((-5)+20))- log (2,((-5)+2)) = log (2,(-5)) }}}, we don't have to go any further, since we already found a negative,  (-5) will not work
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(4), {{{ log (2,(4+20))- log (2,(4+2)) = log (2,4) }}} = {{{ log (2,24)- log (2,(6)) = log (2,4) }}}, there are no negatives, "4" will work
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Here is the graph of the equation
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{{{ graph (500,500,-10,10,-10,10, log ( 2, (x+20)/(x^2+2x))) }}}
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Remember though that you cannot take the log of a negative, your only answer is
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{{{ x = 4 }}}
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Hope I helped, Levi