Question 1012454: Good evening, I have some difficulties with this logarithmic equation:
4log4((x+1)^(1/2)) - log2(5-7x)=0
I tried to solve changing the base in such a way:
4log4((x+1)^(1/2)) = (log4(5-7x))/log4(2)
but I do wrong somewhere solving the arguments.
Can somebody help me explaining the right method?
Many thanks
Found 3 solutions by Boreal, Theo, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! you were on the right track. log 4 (2)=1/2, and dividing by 1/2 is multiplying by 2, which is what I did when I converted to base 2, rather than base 4.
4log4((x+1)^(1/2)) =log2(5-7x)
log 4 (x+1)^2=log 2 (5-7x)
log2 (x+1)^2/log2 (4)=log 2 (5-7x)
log2(4)=2, multiply by 2 both sides
log 2(x+1)^2=log 2 (5-7x)^2
(x+1)^2=(5-7x)^2
x^2+2x+1=25-70x+49x^2
0=24-72x+48x^2
divide both sides by 24
2x^2-3x+1=0
(2x-1)(x-1)=0
x=1/2, 1
check back
4 log 4(2^(1/2)-log 2 (5-7)=0; the second term does not exist.
4 log 4 (3/2)^(1/2)-log2 (5-3.5)=0
log 4 (3/2)^2=log 2 (3/2)
log 2 (3/2)^2/log 2 (4)=log 2 (3/2)
====================
The second term is 0.585 or 2^0.585=1.5
the first term is 4 log 4(3/2)^(1/2)
3/2^(1/2)=1.225
4 log 4 (1.225), and log 4(1.225)=0.1465
0.1465*4=0.586, rounding error.
log 2 (4)=2
multiply through by 2
log 2 (3/2)^2=2 log 2 (3/2)=log 2 (3/2)^2
x=1/2
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i think i have it.
your problem is:
4log4((x+1)^(1/2)) - log2(5-7x) = 0
add log2(5-7x) to both sides of the equation to get:
4log4((x+1)^.5) = log2(5-7x)
you can write it this way because 1/2 = .5
since alog(b) = log(b^a), then:
4log4((x+1)^.5) = log2(5-7x) becomes:
log(((x+1)^.5)^4) = log2(5-7x) which becomes:
log((x+1)^(.5*4)) = log2(5-7x) which becomes:
log4((x+1)^2) = log2(5-7x)
change of base formula can be applied here.
change of base formula says loga(x) = logb(x)/logb(a)
using that formula, we change the base of 4 to base of 2 in the following manner.
log4((x+1)^2) = log2((x+1)^2)/log2(4)
now log2(4) = y if and only if 2^y = 4.
since 2^2 = 4, this means that y is equal to 2 and we get:
log2(4) = 2, and ...,
the formula of log4((x+1)^2) = log2((x+1)^2)/log2(4) becomes:
log4((x+1)^2) = log2((x+1)^2)/2
we now know that log4((x+1)^2) is equivalent to log2((x+1)^2)/2
we go back to our formula of log4((x+1)^2) = log2(5-7x) and replace log4((x+1)^2) with log2((x+1)^2)/2 to get:
log2((x+1)^2)/2 = log2(5-7x)
we multiply both sides of this equation by 2 to get:
log2((x+1)^2) = 2log2(5-7x)
since alog(b) = log(b^a), then log2((x+1)^2) = 2log2(5-7x) becomes:
log2((x+1)^2) = log2((5-7x)^2)
this is true if and only if (x+1)^2 = (5-7x)^2
solving this equation for x, we get x = 1 or x = 1/2.
to confirm whether these solutions are good, we have to go back to the original equation.
the original equation is:
4log4((x+1)^(1/2)) - log2(5-7x) = 0
x = 1 is no good because then log2(5-7x) is the log of a negative number which is not allowed.
the solution has to be x = 1/2 or we have no solution.
x = 1/2 is a valid solution, so all we have to do now is confirm that is makes the original equation a true equation.
you can confirm by replacing x with 1/2 and evaluating the original equation.
you will find that the equation becomes 0 when x = 1/2.
you can also graph the original equation.
you can see that the graph of the equation is equal to 0 when x = .5
that graph is shown below:
Answer by MathTherapy(10552)  (Show Source):
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