Mastering Logarithms: Solving Equations with Logarithmic Functions

[nikita33]

Homework Statement

Im not looking for answers as i know the answers already. i also know that given time and a calculator, i could figure these out, but i have neither for the exams. I am just wondering the general way of setting problems like this up, so i can solve others in the future. my book and prof did not go over this. google hasnt helped and the threads here don't seem to address it either.

log₉[tex]\sqrt{3}[/tex]
log₈1/4
log₃[tex]\sqrt{27}[/tex]

Homework Equations

The Attempt at a Solution

9^x=[tex]\sqrt{3}[/tex]
8^x=1/4
3^x=[tex]\sqrt{27}[/tex]
? is there an easier or more logical way? these equations are not helping me.
i still do not know how to figure out what x is. even if i memorise the exponents, i think there must be an actual method to do these.

 

[BoundByAxioms]

Try the change of base formula:

log_ba = [tex]\frac{log(a)}{log(b)}[/tex]

 

[nikita33]

ok, so if i do, for example,

i feel i am still in the same predicament of not knowing how to find out what the exponent (to any base i choose) would be (meaning 3^x=[tex]\sqrt{27}[/tex], how would i figure out the x).
eg., if i were to use base 3 for both, log₃3 is 1 but i still am left with log₃[tex]\sqrt{27}[/tex] which i don't know how to figure out, since i can't use the calculator and i don't think memorizing will help me in the future.

i guess, for the above problem, the only way to do it is to put 27 under the radical and take the 3 root? of course, I've been thinking about it for a few days, but i just had a quiz that had a problem like this and i didnt answer it because the quiz was 10 minutes and i didnt have enough time to tinker with it. plus, i don't see how it will work with log₈1/4.

of course, i could be completely off base here, as i just started precalc and am in the learning process. sorry for babbling on. thank you though.

 

[HallsofIvy]

Using that formula is not the best way to do this problem. The answers follow directly from the definition of the logarithm and using basic definitions and facts is always better than just applying formulas (true, it involves thinking which is often painful!)

You are starting well when you change to exponential form:
[itex]9^x=\sqrt{3}[/itex]
[itex]8^x=1/4[/itex]
[itex]3^x= \sqrt{27}[/itex]

Now write the right side as an exponential also: [itex]\sqrt{27}= 3^?[/itex].

For the first and second, it will probably be easiest if you have the same base for both sides. 9= 3² so [itex]9^x= (3^2)^x= 3^?[/itex]. [itex]\sqrt{3}= 3^?[/itex].

8= 2³ so [itex]8^x= (2^3)^x= 2^?[/itex]. 4= 2² so 1/4= 2^?.

 

[nikita33]

youre right about the thinking hurting. sometimes i feel like my brain is at terminal capacity, but i hate having these open questions.

ive figured out, albeit backwards, that 1/4 is 2^-2 so
2^3x=2^-2
base 2 cancels leaving 3x=-2 and algebra = 8^-2/3=1/4. it seems so obvious, and i did figure it out a few days ago, but i needed a systematic way of doing it.

i did log₉[tex]\sqrt{3}[/tex] by
3^2x=3^1/2
base 3 cancels, leaving 2x=1/2, 4x=1 and its 9^1/4=[tex]\sqrt{3}[/tex]

finally, 3^x=3^31/2
base 3 cancels, x=3/2

so, i guess the moral of the story is to try to get like bases. i just learned logs for the first time in my life, so its kind of like a new language, even though i understand exponents. i think i just need to get used to the idea.

thanks so much for laying out the path for me to solve these.