Strategies for Solving Logarithmic Find X Equations

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ok I am having abit of trouble solving the find x type of questions
the 3 questions I've got are

1.) solve x for the following equation
loge(x) + loge(3x+1) = 1

2.) solve x
2loge(x) + loge4 = loge(9x-2)

3.) Given that loga(N) = 1/2(loga(24) - loga(0.375) - 6loga(3) ) find the value of N ..now i was wondering if ne-one would be able to help me with these types of questions cause i don't know how to approach em:bugeye:
 
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Are you familiar with logarithmic identities?

log(a)+log(b)=log(ab)
log(a)-log(b)=log(\frac{a}{b})
b\cdot log(a)=log(a^b)
 
You must have been studying the properties of logs.

What is logx + logy equivenent to?

What about alogx ? Look at your basic properties and think about how to apply them to your problem.
 
Yeah i am familiar with them and when i tried applying them i got answers that didnt match when i tried putting them in a ti-83 calculator to match u can use second table to get rough answers but i would appreciate being bale to do them without a calculator
 
Show us your work.
 
oh its alright I've managed to figure out 2 and 3 after extensive rubbing out and many attempts lol now I am sure if i can do them that i should be able to do the first one
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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