Question 587477: use change of base formula to solve 2^2x = 90
Answer by xdragonfight(116)   (Show Source):
You can put this solution on YOUR website! Solve 2x = 30.
If this equation had asked me to "Solve 2x = 32", it would have been easy, because I could have converted the 32 to 25, set the exponents equal, and solved for "x = 5". But 30 is not a power of 2, so I can't set powers equal to each other. I need some other method of getting at the x, because I can't solve with the equation with the variable floating up there above the 2; I need it back down on the ground where it belongs. And I'll have to use logarithms to get at it.
When dealing with equations, I can do whatever I like to the equation, as long as I do the same thing to both sides. And, to solve an equation, I have to get the variable by itself on one side of the "equals" sign; to isolate the variable, I have to "undo" whatever has been done to it.
In this case, the variable x has been put in the exponent. The backwards (technically, the "inverse") of exponentials are logarithms, so I'll need to undo the exponent by taking the log of both sides of the equation. This is useful to me because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log:
logb(mn) = n · logb(m)
When I take the log of both sides of an equation, I can use any log I like (base-10 log, base-2 log, natural log, etc), but some are sometimes more useful than others. Since the base in the equation "2x = 30" is "2", I might try log-base-2:
log2(2x) = log2(30)
xlog2(2) = log2(30)
x(1) = log2(30)
x = log2(30)
But I can't evaluate this in my calculator unless I apply the change-of-base formula:
x = log2(30)
= ln(30)/ln(2)
What would happen if I just used the natural log in the first place?
2x = 30
ln(2x) = ln(30)
xln(2) = ln(30)
x = ln(30)/ln(2)
Either way, I get the same answer, but taking natural log in the first place was simpler and shorter.
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