Question

Solve 3 log6 2 + 2 log6 3 − log6 x = 2.

Answer 1

a.x = 2
b.x = 6
c.x = 15
d.x = 36

Answer 2

This is where all the log rules come into play.

Answer 3

yes, logarithms give me a headache

Answer 4

some rules of logarithms:

\(\color{#000000 }{ \displaystyle \log_a(b)+\log_a(c)=\log_a(b\cdot c) }\)

\(\color{#000000 }{ \displaystyle \log_a(b)-\log_a(c)=\log_a(b\div c) }\)

\(\color{#000000 }{ \displaystyle w\log_a(b)=\log_a(b^w) }\)

Answer 5

would it be x=6?

Answer 6

\(\color{#000000 }{ \displaystyle 3\log_6(2)+2\log_6(3)-\log_6x=2 }\)

write it as a single logarithm and then use (another rule)
\(\color{#000000 }{ \displaystyle \log_a(b)=c\quad \Longleftrightarrow \quad a^c=b }\)

Answer 7

Use the rules of logs on the left side to end up with a single log of a number.

Answer 8

You want an example, perhaps?

Answer 9

yes please

Answer 10

\(\color{#0000ff }{ \displaystyle 4\log_2(2)-2\log_2(x)+ 2\log_2(10)=2}\)

after applying; \(w\log_a(b)=\log_a(b^w)\), we get:

\(\color{#0000ff }{ \displaystyle \log_2(2^4)-\log_2(x^2)+ \log_2(10^2)=2}\)
\(\color{#0000ff }{ \displaystyle \log_2(16)-\log_2(x^2)+ \log_2(100)=2}\)

then, we'll use; \(\log_a(B)+\log_a(C)=\log_a(B\cdot C)\),
and \(\log_a(B)+\log_a(C)=\log_a(B\cdot C)\) we get:

\(\color{#0000ff }{ \displaystyle \log_2(16\div x^2\times 100)=2}\)

or, \(\color{#0000ff }{ \displaystyle \log_2\left(\frac{1600}{x^2}\right)=2}\)

And then we apply; \(\color{ }{ \displaystyle \log_a(b)=c\quad \Longleftrightarrow \quad a^c=b }\)

\(\color{#0000ff }{ \displaystyle \left(\frac{1600}{x^2}\right)=2^{2}}\)

and then it's just a algebra

\(\color{#0000ff }{ \displaystyle \frac{1600}{x^2}=4}\)

\(\color{#0000ff }{ \displaystyle \frac{1600}{4x^2}=1}\)

\(\color{#0000ff }{ \displaystyle 1600=4x^2 }\)
\(\color{#0000ff }{ \displaystyle 400=x^2\quad \Longrightarrow \quad x=\pm20 }\)

Answer 11

I am so confused :(