Logarithms: log_3(1/8), log_2(1/128), log_36(6), etc

[Hockeyman]

Have just a few questions with Logarithms before my test tommorow.

Log {base3} (1/81)

How would you get the answer for this? I tried (1/3)^4, but (1/3) still isn't the same as 3 so how would you solve this?

Log {base2} (1/128)

Same problem here can't figure out how to turn the (1/128) to have a base of 2

Log {base36} 6

Same problem.


This last one is alittle different :

Log {base10} y=(1/4) Log {base10} 16 + (1/2) Base {base10} 49

Using the Product Property and Power Property i can simplify alittle

Log {base10} y=(1/4) Log {base10} 112

I'm not sure where to go from here?

 

[stapel]

1) log₃(1/8)

What are the instructions? (Since there is no equation here, there is nothing to "solve".)

2) log₂(1/128)

If the instructions for this are to "simplify", then think about the power of 2 that equals 1/128. (Hint: Think about negative exponents.)

3) log₃₆

If the instructions for this are to "simplify", then think in terms of roots-as-exponents.

4) log₁₀(y) = (1/4)log₁₀(16) + (1/2)log₁₀(49)

If the instructions for this are to "solve", then use one of the log rules to take the multipliers inside as powers. For instance, (1/4)log₁₀(16) = log₁₀(16^1/4). Then combine the two longs on the right-hand side into one (something you can only do after the multipliers are out of the way).

Once you have "log(something) = log(something else)", you can set "something" equal to "something else", and solve the resulting equation.

Eliz.

 

[Hockeyman]

Have just a few questions with Logarithms before my test tommorow.

Evaluate each expression.

Log {base3} (1/81)

So that makes this 3^-4 = -4


Log {base2} (1/128)

This 2^-7 = -7

Log {base36} 6

And this 36^(1/2) = (1/2)

Correct?

The last one i now understand.

 

[soroban]

Hello, Hockeyman!

Here's one way.
it takes an extra step, but it should make things clearer for you.

\(\displaystyle \log_3\left(\frac{1}{81}\right)\)

Let \(\displaystyle \log_3\left(\frac{1}{81}\right)\;=\;x\)

Rewrite in exponential form: \(\displaystyle \,3^x\:=\:\frac{1}{81}\)

Get the same base on both sides: \(\displaystyle \,3^x\:=\:\frac{1}{3^4}\)

\(\displaystyle \;\;\) and we have: \(\displaystyle \,3^x\:=\:3^{-4}\)

We have the same base on both sides.
\(\displaystyle \;\;\)Therefore, the exponents are equal: \(\displaystyle \;x\:=\:-4\)

Therefore: \(\displaystyle \,\log_3\left(\frac{1}{81}\right)\:=\:-4\)

\(\displaystyle \log_2\left(\frac{1}{128}\right)\)

We have: \(\displaystyle \,\log_2\left(\frac{1}{128}\right)\:=\:x\;\;\Rightarrow\;\;2^x\:=\:\frac{1}{128}\:=\:\frac{1}{2^7}\)

Then: \(\displaystyle \,2^x\:=\:2^{-7}\;\;\Rightarrow\;\;x\:=\:-7\)

Therefore: \(\displaystyle \,\log_2\left(\frac{1}{128}\right)\:=\:-7\)

\(\displaystyle \log_{36}6\)

We have: \(\displaystyle \,\log_{36}6\:=\:x\;\;\Rightarrow\;\;36^x\:=\:6\)

We need to get 36 on both sides: \(\displaystyle \,36^x\;=\;36^{\frac{1}{2}}\;\;\Rightarrow\;\;x\:=\:\frac{1}{2}\)

Therefore: \(\displaystyle \,\log_{36}6\:=\:\frac{1}{2}\)

\(\displaystyle \log_{10}y\:=\:\frac{1}{4}\log_{10}16\,+\,\frac{1}{2}\log_{10}49\)

Your work was close . . .

We have: \(\displaystyle \;\log_{10}y\;=\;\log_{10}\left(16^{\frac{1}{4}}\right)\,+\,\log_{10}\left(49^{\frac{1}{2}}\right) \;=\;\log_{10}(2)\,+\,\log_{10}(7) \;= \;\log_{10}(2\cdot7)\)

Hence, we have: \(\displaystyle \:\log_{10}y \:=\:\log_{10}14\)

"Un-log" both sides: \(\displaystyle \:y\:=\:14\)