Logarithim Help

[PuppyMilk]

Hi, I'm new here and not sure whether this subject would fit in this topic, so apologies if I'm off to a bad start.

I wouldn't usually have trouble with this, but it has been a while since I've done this topic and I've gone completely blank.

I have only a couple of questions, so any help would be appreciated.

The first one I need to solve algebraically:

2ⁿ⁺¹ = 1/32

and another two I was having a problem with:

Given that log_p7 + log_pk = 0, find k

and

Given that 4log_q3 + 2log_q2 - log_q144 = 2. find q.

If you could give me a little bit of help when finding the answer, that'd be great. Thank you in advance.

 

[pappus]

Hi, I'm new here and not sure whether this subject would fit in this topic, so apologies if I'm off to a bad start.

I wouldn't usually have trouble with this, but it has been a while since I've done this topic and I've gone completely blank.

I have only a couple of questions, so any help would be appreciated.

The first one I need to solve algebraically:

2ⁿ⁺¹ = 1/32

and another two I was having a problem with:

Given that log_p7 + log_pk = 0, find k

and

Given that 4log_q3 + 2log_q2 - log_q144 = 2. find q.

If you could give me a little bit of help when finding the answer, that'd be great. Thank you in advance.

1. If you could remember that \(\displaystyle \displaystyle{32 = 2^5}\) then the 1st equation can be solved quite easily:

\(\displaystyle \displaystyle{2^{n+1}= \frac1{32}~\implies~2^{n+1}= \frac1{2^5}~\implies~2^{n+1}= 2^{-5}}\)

Two powers with the same base are equal if the exponents are equal too. Use this property and solve for n.

 

[pappus]

...

and another two I was having a problem with:

Given that log_p7 + log_pk = 0, find k

and

Given that 4log_q3 + 2log_q2 - log_q144 = 2. find q.

If you could give me a little bit of help when finding the answer, that'd be great. Thank you in advance.

Here are some logarithm rules you should know:

\(\displaystyle \displaystyle{\log_b(1) = 0}\)

\(\displaystyle \displaystyle{\log_b(x) + \log_b(y) = \log_b(x \cdot y)}\)

\(\displaystyle \displaystyle{\log_b(x^y) = y \cdot \log_b(x)}\)

\(\displaystyle \displaystyle{\log_b(x) - \log_b(y)=\log_b\left(\frac xy \right)}\)

\(\displaystyle \displaystyle{b^{\log_b(x)} = x}\)