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 Post subject: Two problems, please help!Posted: Wed, 29 Feb 2012 03:31:45 UTC
 S.O.S. Newbie

Joined: Wed, 29 Feb 2012 03:23:58 UTC
Posts: 4
I have two fairly simple problems, and I know I could probably find the answers, but I want to understand the steps so I can complete future problems like these myself..If you know what you're doing please give me the steps and the answer would help too

Solve:
1.6^(4x)=102
2.log[base3]+log(4x-1)=5

Any help would be greatly appreciated!

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 03:56:22 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
cayliejohnson369 wrote:
I have two fairly simple problems, and I know I could probably find the answers, but I want to understand the steps so I can complete future problems like these myself..If you know what you're doing please give me the steps and the answer would help too

Solve:
1.6^(4x)=102
2.log[base3]+log(4x-1)=5

Any help would be greatly appreciated!

If you want help on something you can do yourself you should at least post some work to show you care enough to try.

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 04:00:51 UTC
 S.O.S. Newbie

Joined: Wed, 29 Feb 2012 03:23:58 UTC
Posts: 4
That doesn't help me at all.
And I have tried to work on the problems..but I've come up with answers that I don't think are right plus I don't know the correct steps to take. How about only posting when you have an answer or steps..not to judge.

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 04:06:04 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
cayliejohnson369 wrote:
That doesn't help me at all.
And I have tried to work on the problems..but I've come up with answers that I don't think are right plus I don't know the correct steps to take. How about only posting when you have an answer or steps..not to judge.

I'm not here to judge, but this board is to help people who are having trouble, not those who just don't want to do their own work. If you have some candidate answers you should provide them along with what you did go get them, this shows us you're at least invested enough in getting the answer to try for it yourself. Our time is also valuable, I'm just letting you know outright that you're wasting your time here if you expect answers without any effort on your part, most users will just ignore questions like this without letting the poster know that they could be using their time better elsewhere.

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 18:01:32 UTC
 Member

Joined: Tue, 20 Oct 2009 18:18:45 UTC
Posts: 33
Location: Lisbon, Portugal
following shadow thoughts, i will give you two clues:

solving a exponential equation, you will need, at the final, to use the inverse: a logarithm with the same basis

we have a^(x-y)=(a^x)/(a^y), so you use this property: Log (x/y) = Log x - Log y

if you have some scenarios of your work, show them and i help you finding your mistakes

Voxx

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 18:24:23 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3693
Location: Ottawa Ontario
cayliejohnson369 wrote:
1.6^(4x)=102

If a^x = b, then x = log(b) / log(a)

Since you're apparently unaware of this basic rule, then not much
can be done for you at a "help" site ; not a classroom.

If you were helping out at a help site for "bicycle riding",
what would you answer to: "where are the pedals"?

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 Post subject: Re: Two problems, please help!Posted: Wed, 29 Feb 2012 19:04:57 UTC
 S.O.S. Oldtimer

Joined: Sat, 16 Aug 2008 04:47:19 UTC
Posts: 208
cayliejohnson369 wrote:
I have two fairly simple problems, and I know I could probably find the answers, but I want to understand the steps so I can complete future problems like these myself..If you know what you're doing please give me the steps and the answer would help too

Solve:
1.6^(4x)=102
2.log[base3]+log(4x-1)=5

Any help would be greatly appreciated!

I don't understand what number 2 means..

??

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