Question 247100
graph of equation as y = 2*log(3,(x+4))


{{{graph (600,600,-10,10,-10,10,2*log(3,(x+4)))}}}


graph of equation as y = log(3,(x+4)^2))


{{{graph (600,600,-10,10,-10,10,log(3,((x+4)^2)))}}}


As long as x+4 is positive, the graphs are identical, as they should be.


The wrinkle comes in when x+4 is zero or negative.


The first graph doesn't plot for (x+4) is zero or negative because you can't take the log of 0 or a negative number.


(x+4) becomes negative when x is less than or equal to -4.


The second graph plots for (x+4) is 0 or negative, because (x+4)^2 becomes positive regardless of whether the expression (x+4) is positive or negative.


This winds up being a curious anomaly because the laws of logarithms state that:


log(x^y) = y*log(x) which is essentially what I did in reverse.


I did y*log(x) = log(x^y).


They are equal as long as the expression is positive.


This situation shows that it is not when the expression is zero or negative


When (x+4) is positive, the graphs are identical.


When (x+4) is zero or negative, we have one equation providing values while the other equation says no can do.


Regardless of that anomaly, we can still find the equivalent exponential expression.


your equation starts out as:


y = 2*log(3,(x+4))


This translates to y = 2 * log of x to the base 3.


since e*log(b,x) = log(b,x^e), your equation becomes


y = log(3,(x+4)^2) which is equivalent to y = 2*log(3,(x+4))


The basic laws of logarithms states that:


y = log(b,x) if and only if b^y = x


In your equation that becomes:


y = log(3,(x+4)^2) if and only if 3^y = (x+4)^2


This equation is the same as:


(x+4)^2 = 3^y


Take the square root of both sides of this equation to get:


x+4 = sqrt(3^y)


subtract 4 from both sides of this equation to get:


x = sqrt(3^y) - 4


If you had done what you were recommended to do, you would have wound up with the same answer.


The recommendation was to divide both sides of the equation by 2.


Your equation was:


y = 2 * log(3,(x+4))


divide both sides of this equation by 2 to get:


y/2 = log(3,(x+4))


This is true if and only if 3^(y/2) = x+4


this is the same as:


x + 4 = 3^(y/2)


subtract 4 from both sides of this equation to get:


x = 3^(y/2) - 4


since 3^(y/2) = 3^(y*(1/2)) = (3^y)^(1/2) = sqrt(3^y), then:


your equation becomes:


x = sqrt(3^y) - 4


which is the same as I derived above.