The first type of logarithmic equation has two logs, each having the same base, which have been set equal to each other. We solve this sort of equation by setting the insides (that is, setting the "arguments") of the logarithmic expressions equal to each other. For example:
The logarithms on either side of the equation have the same base;namely, a base of 2. The only way these two log expressions can be equal is for their arguments to be equal. In other words, the log expressions being equal says that the arguments must be equal, so I can create the following equation:
x = 14
Content Continues Below
And that's all there is to solving this equation. My solution is:
x = 14
The base of these logarithmic terms is unknown, being indicated by the letter b. But that's okay. I only need them bases to be the same. What those bases actually are doesn't matter for this sort of equation. Because the bases of the logs are the same, then I know that the insides of the logs must be equal. I'll use this to create my equation:
x^{2} = 2x – 1
Then I can solve the log equation by solving this quadratic equation:
x^{2} – 2x + 1 = 0
(x – 1)(x – 1) = 0
Then my solution is:
x = 1
Affiliate
Affiliate
Logarithms cannot have non-positive arguments (that is, arguments which are negative or zero), but quadratics and other equations can have negative solutions. When we convert a log equation to a different type of equation by equating the insides of the logs, we may be "creating" solutions that didn't previously exist. Because of this, it is generally a good idea to check the solutions you get for log equations.
To check my solution for the exercise above, I'll plug my solution value of x = 1 into each side of the equation, and see if I get the same value for each side:
the left-hand side:
log_{b}(x^{2}) = log_{b}(1^{2}) = log_{b}(1) = 0
the right-hand side:
log_{b}(2x – 1) = log_{b}(2(1) – 1) = log_{b}(2 – 1) = log_{b}(1) = 0
The fact that each side of the original equation evaluated to the same value (in this case, to the value of zero) proves that my solution is correct.
It should be noted that the particular value of the base of the log was irrelevant here. Each log in the equation had the same base, and each side of the log equation ended up with the value, so the solution "checks".
Since the logs have the same base, I can set the arguments equal and solve:
x^{2} – 30 = x
x^{2} – x – 30 = 0
(x – 6)(x + 5) = 0
x = 6, – 5
Since I cannot have a negative inside a logarithm, the quadratic-equation solution "x = –5" can not be a valid solution to the original logarithmic equation (in particular, this negative value won't work in the right-hand side of the original equation).
Then my solution is:
x = 6
Content Continues Below
All of these logs have the same base, but I can't solve yet, because I don't yet have the equation in the form "log(of something) equals log(of something else)". So first I'll have to apply some log rules to condense the log expression on the right-hand side of the equation and to move the multiplier on the left-hand side to inside that log:
2log_{b}(x) = log_{b}(4) + log_{b}(x – 1)
log_{b}(x^{2}) = log_{b}((4)(x – 1))
log_{b}(x^{2}) = log_{b}(4x – 4)
Now that I've rearranged the original equation to put it in the proper "log(of something) equals log(of something else)" form, I can equate the logs' arguments and solve the resulting equation:
x^{2} = 4x – 4
x^{2} – 4x + 4 = 0
(x – 2)(x – 2) = 0
Then my solution is:
x = 2
Affiliate
Advertisement
To solve this, I need to remember the defintion of logarithms. Logarithms are powers. Specifically, "log_{b}(a)" is the power that, when put on the base "b", gives me "a". In this case, the base of the log is e. The argument of "ln(e^{x})" is "e^{x}". That is, "ln(e^{x})" is "the power that, when put on e, gives you e^{x}.
What power do I have to put on e to get e^{x}? Why, x, of course! So:
ln(e^{x}) = x
Similarly:
ln(e^{3}) = 3
...and:
ln(e^{5}) = 5
So the given equation simplifies quite nicely:
ln(e^{x}) = ln(e^{3}) + ln(e^{5})
x = 3 + 5
x = 8
Then my solution is:
x = 8
Some students like to think of the statement "log_{b}(b^{x}) = x" as a "cancellation" of some sort. Technically, that isn't a correct statement. But this type of expression often proves quite confusing. If thinking of this as a "cancellation" helps you to keep things straight in your mind, then please go right ahead. Just don't use that lingo in class, or your instructor might get a bit tetchy.
Note: The above equation could also have been solved by using log rules:
ln(e^{x}) = ln(e^{3}) + ln(e^{5})
ln(e^{x}) = ln[(e^{3})(e^{5})]
ln(e^{x}) = ln(e^{3 + 5})
ln(e^{x}) = ln(e^{8})
Comparing the arguments, I get:
e^{x} = e^{8}
x = 8
These log equations were fairly simple to solve, because of their form; namely, "log(of something) equals log(of something else)". But if a given log equation is not in this form, and cannot be rearranged to be put into this form, then we'll have to bring exponentials into the mix...
URL: https://www.purplemath.com/modules/solvelog.htm
© 2018 Purplemath. All right reserved. Web Design by