Question 1095930: How do you solve using logarithms when you can't express with same bases?
Question: 5^(2x+1) = 2^(x-2)
Thankyou!
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39630)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! there are a couple of ways.
your problem is:
5^(2x+1) = 2^(x-2)
first way is to take the log of both sides of this equation to get:
log(5^(2x+1)) = log(2^(x-2))
log by itself is equal to log base 10, sometimes shown as log10.
if you see log(x) by itself then assume it is really log10(x).
log10 is the log function on your calculator.
by the basic properties of logarithms:
log(x^a) = a*log(x)
therefore:
log(5^(2x+1)) = (2x+1) * log(5) and log(2^(x-1)) = (x-1) * log(2)
your equation becomes:
(2x+1) * log(5) = (x-2) * log(2) ***** note 1
***** note 1
this equation will be referenced again below when we get to the second method.
***** end of note 1
simplify to get:
2x * log(5) + 1 * log(5) = x * log(2) - 2 * log(2)
subtract x * log(2) from both sides of the equation and subtract 1 * log(5) from both sides of the equation to get:
2x * log(5) - x * log(2) = -2 * log(2) - 1 * log(5)
factor out the x to get:
x * (2 * log(5) - 1 * log(2)) = -2 * log(2) - 1 * log(5)
divide both sides of this equation by (2 * log(5) - 1 * log(2)) to get:
x = (-2 * log(2) - 1 * log(5)) / (2 * log(5) - 1 * log(2))
remember that log(x) = log10(x) = LOG function on your calculator.
use your calculator to solve for x to get:
x = -1.186086352
that should be your answer.
replace x in the original equation with that to get:
5^(2x+1) = 2^(x-2) becomes .1098733685 = .1098733685.
the solution is confirmed to be correct.
you can also solve it as follows:
this is the second way and actually winds up being the same as the first way as you will see.
start with 5^(2x+1) = 2^(x-2)
by the basic definition of logs:
5^(2x+1) = a if and only if log5(a) = 2x+1
2^(x-2) = b if and only if log2(b) = x-2
since you are given that 5^(2x+1) = 2^(x-2), then it follows that a must be equal to b.
therefore you get:
5^(2x+1) = a if and only if log5(a) = 2x+1
and you get:
2^(x-2) = a if and only if log2(a) = x-2
there is a base conversion formula that says:
loga(x) = logb(x) / logb(a)
in english, that formula states:
log to the base a of x = log to the base b of x divided by log to the base b of a.
specific to your problem, you have:
log5(a) = 2x+1 becomes log(a) / log(5) = 2x+1
and:
log2(a) = x-2 becomes log(a) / log(2) = x-2
if you multiply the first of these equations by log(5), you get log(a) = (2x+1) * log(5)
if you multiply the second of these equations by log(2), you get log(a) = (x-2) * log(2)
since they're both equal to log(a), then you get:
(2x+1) * log(5) = (x-2) * log(2) ***** note 2
***** note 2
this is the same formula we derived above using the first method
see ***** note 1 above.
***** end of note 2
as you can see, both methods lead to the same equation.
it is far simpler to just take the log of both sides of the equation as we did in the first method.
if you decided to use the second method, you would wind up with the same equation as in the first method which would lead you to the same solution.
here's some good references for you that can help.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/
tutorials 26 and 29 deal with exponents
http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/index.htm
tutorials 23 and 24 deal with exponents and scientific notation.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/
tutorials 42 through 48 deal with exponent and log equations and how to solve them.
this is the one that has the meat of the topic in it.
if you already know the basics, then go here.
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