Question 35520: how do you solve equations that have different logs: log5(x-4) = log7x, solve for x
The 5 and the 7 are little Answer by narayaba(40) (Show Source):
You can put this solution on YOUR website! This problem cannot solved by mere analysis it requires that one needs to find a numerical solution for this problem. But here I am going to provide a graphical solution.
Since the logarithms on both sides are to the different bases. First we need to bring to them the same base.
log5(x-4) = log7x (equiation 1)
log of a to b can be written as log of a to base c divided by log of b to base c
which means logb(a) = logc(a)/logc(b)
The right hand side of equation 1 can be rewritten as log5(x)/log5(7) according to the above mentioned procedure.
thus log5(x-4) = log5(x)/log5(7)
this is same as log5(x) = log5(7) * log5(x-4)
log5(7) is 1.2091
log5(x) = 1.2091 * log5(x-4)
log5(x) = log5((x-4)^1.2091) Note: m*loga(b) = loga(b^m)
since the lograthims on both sides are to the same base
x = (x-4)^1.2091
now we need to solve for x. It is not easy to solve for x as 1.2091 is not a perfect integer
In order to solve this
we write y = x - (x-4)^1.2091
Now we graph y as a function of x for different values of x
From the graph we see the value of x at which y becomes 0
the value of x at which y becomes 0 is 11.282
we can verify this value by plugging into the original equation
log5(x-4) = log7x
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