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Jim Fisher Posted on Feb 11, 2014
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kakima

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  • Netflix Master 102,366 Answers
  • Posted on Feb 12, 2014
kakima
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0helpful
1answer

2log(base4)x + log(base4)3=log(base4)x-log(base4)2

Hi,
It is easier to use the rules for logarithms to clean this equation before trying to solve it. And the calculator is no help here.
Let me represent the logs as follows log b in base a = log(a,b). The first argument is the base.

2*log(4,x) - log(4,3)= log(4,x)- log(4,2)

Here you have two similar terms 2*log(4,x) on the left and log(4,x) on the right. Gather them both on the same side, say the left. You have

2*log(4,x) -log(4,x) +log(4,3) = log(4,x)-log(4,x) -log(4,2)

which is equivalent to log(4,x) +log(4,3) = -log(4,2)

Gather the constant terms on the right, and use log(a) +log(b)=log(a*b)

log(4,x) = -log(4,2) -log(4,3) = - { log(4,2)+log(4,3)} = -log(4,2*3)

You get log(4,x)= -log(4,6).

Take the constant term to the left, while changing its sign in the process. On the right remains 0, which is also the log of 1 in any base So log(4,x)+log(4,6) =0 =log(4,1)

Use the rule log(a) +log(b) = log(a*b) valid in any base, to obtain
log(4,6x)=log(4,1) which is equivalent to 6x=1. The solution is x=1/6

There are a few other shortcuts to get the solution but I preferred to take a pedestrian, step by step, approach.

Hope it helps.


10helpful
2answers

I need help with a change of base formula problem.

It's simple here some logs:

log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
you just need to know that the default base of log in calculators is 10, so log x mean
log x base 10 = log x / log 10,
so write log x / log y for log x base y example:
log 8 base 2 = log 8 / log 2 that is how you write it on calculator, hope it was helpful
7helpful
2answers

How do you input a logarithm base into the ti83 plus?

You can't it's default 10 so learn this logs:
log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
so just divide the logs
hope it helps
4helpful
1answer

How to calculate

It's very simple: the calculators have a default base 10 so that log x it's log x base 10 so here is some logs:
log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
Hope it's helped
0helpful
1answer

To solve the problem using log

Look the calculators has a default base 10 and you can't change it but you may learn this:
log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
so instead input log 8 base 2 you need to input log 8 / log 2
your example answer is about 0.919... if i get right what did you write.
Glad to help :)
0helpful
1answer

Find log(1-.2)

I don't get it but here some help with logs:
log x base y = log x / log y (when log x is default base 10 in calculators) and
log (x*y) = log x + log y
log (x/y) = log x - log y
the bases is default 10
hope it's helped
2helpful
1answer

Logarithmic functions

All calculators it's the same:
log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y
the default base in all the calculators is 10 so instead of writing log 4 base 2 you simply need to write log 4 / log 2
Hope you understand :)
1helpful
1answer

Using logs

Here something that would help you:

log x base y = log x / log y
log (x*y) = log x + log y
log (x/y) = log x - log y

so log 24 base 2 it's log 24 / log 2 etc.
and if i get right what you wrote x= 10 ^ ( 3 * (log 5) ) if no inverse it it's easy really Hope it's helped you.
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