Can you help me with these logarithm problems?

Question

Hi, I am desperately needing some help with these equations.
My goal is to find x and I'm a little lost.
I hope I can explain these well enough that someone knows what I'm talking about.

The first one is:
log3x=2

The second problem looks like:
-6^-3x=-73
(*Note: the '^' is "to the power of"...So it'd be like -6 to the power of -3x...I hope that makes sense!)

Third:
2logbase7 x - logbase7 7=logbase7 7
(This one was really hard to explain. The two is out in front of the log, followed by a base 7 and then an x..minus a log base 7 followed by just a 7...equals..a log base 7, followed by a regular 7...I put a space after the base to indicate that it's not part of the base...Does that make sense?)

The next one is:
logbasen 4=5
(So log base 'n' followed by a 4...equals 5..)

Next is:
logbase5 (x-4)=3
(Okay...a log with a base 5 followed by (x-4) equals 3.)


If anyone could help me out with these problems I would be extremely thankful. I've been struggling to solve these all day.

Answer 1

1) I'm going to assume that you mean
log(3x) = 2

The fact that there's no base means that it is log base 10. {Other may disagree with me and believe that it means log base e, but log base e has its own function, ln}.

To change from logarithmic to exponential form, note the following:
log[base b](a) = c if and only if b^c = a.

Therefore

10^2 = 3x.
100 = 3x. Divide both sides by 3,

100/3 = x

2) -6^(-3x) = -73

Bases of logs are usually not negative, so I'm presuming the negative sign in front of the 6 is not part of the base. That is, I'm assuming -6^(-3x) instead of (-6)^(-3x).
With that said, this means we can multiply both sides by -1

6^(-3x) = 73

Changing this to logarithmic form,

log[base 6](73) = -3x. Dividing both sides by (-3),

(-1/3) log[base 6](73) = x

3) 2log[base 7](x) - log[base 7](7) = log[base 7](7)

Adding log[base 7] to both sides, we can essentially "group like terms" of the log[base 7](7) on both sides.

2log[base 7](x) = 2log[base 7](7)

Dividing both sides by 2,

log[base 7](x) = log[base 7](7)

And now, we can just equate the insides of the logs, and
x = 7.

4) log[base n](4) = 5

To solve this, change into exponential form (refer to above to know how to do it). That means

n^5 = 4. Therefore, presuming we only want real solutions (as there are 4 other complex solutions)

n = 4^(1/5)

5) log[base 5](x - 4) = 3

Based on these questions, all I can conclude you brush up on is how to change from logarithmic form to exponential form, and vice versa. A reminder again, that
log[base b](a) = c is the same as b^c = a

5^3 = x - 4
125 = x - 4
129 = x

Answer 2

I´ll suppose that if you don´t mention any base the base is 10
1) by definition of logarithm
if log 3x = 2 that means that 3x =10^2 so x= 100/3

2) 6^-3x =73 Take log of bothsides and apply known rules
-3x*log6 = log 73 so x = - log73/(3log6) = -0,7982

3)Now we are working in base 7

Remember that the log of the base is 1 So
logx^2 -1 = 1 log x^2 = 2 and x^2 =10^2 = 100 so x= 10. As negative numbers don´t have log x=-10 can´t be

4) If log in base n of 4 is 5 that means that n^5 = 4

Taking log in base 10 of both sides 5logn =log4 so

n= log4/5 =0.1204

5) x-4 =5^3 so x= 4+125 =129

OK I made your homework.Next time study and do it by yourself

Answer 3

log3x = 2
assuming base 10 logs 3x=10^2 by definition
therefore 3x=100 and x = 33.33333.........

-6^-3x = -73 can be written as
- 1/6^3x = -73
simplifying we get 1 = 73(6^3x) or 1/73 =6^3x
now take logs of both sides: log(1/73) =3x log(6)
or [log (1) -log (73)] / [3log(6)] = x
now look the numbers up in a base 10 log table and do the arithmatic.

2 logbas7( x) - logbase7(7) =logbas7(7)
add logbas7(7) to both sides
2logbas7(x) =2logbas7(7)
obviously x=7

logbasen (4) = 5
by definition of log: n^5 =4
then taking base 10 logs of both sides
5*log10(n) =log10(4)
log10(n) =log10(4)/5
now look up log 10 of 4, divide it by 5 and look up the antilog.

logbas5(x-4) = 3
(x-4) =5^3 = 125
x = 125/4 =31.25

Answer 4

1. lg3x = 2
3x = 10^2
x = 100/3
=33.3

2. -6^-3x = - 72
6^-3x = 72
lg(6^-3x) = lg72
(-3x)lg6 = lg 72
x = -lg72/3lg6
= -0.796

3. 2logbase7 x - logbase7 7 = logbase7 7
2logbase7 (x^2) = 2logbase7 7
x^2 = 1
x = -1 (n.a) or 1

4. logbasn 4 = 5
4 = n^5
n = 4^(1/5)
= 1.32

5. logbase5 (x - 4) = 3
x - 4 = 5^3
x = 125 + 4
= 129

Logarithms are actually very easy to solve - you just need to know the logarithmic laws (e.g. lg (ab) = lg a + lg b) and apply them to the questions.