Question

(log_(3) (log_(3) x)) = -1 solve for x

Answer 1

\(\Large log_3[log_3(x)]=-1\quad ?\)

Answer 2

yes

Answer 3

do you know how to convert log notation to exponential notation?

Answer 4

kind of this one would be 3^-1 = 1/3 so log_3(1/3) ?

Answer 5

because a^b=c and B^c=a

Answer 6

well, more or less, yes \(\bf log_3[log_3(x)]=-1\implies 3^{-1}=log_3(x)\implies \cfrac{1}{3}=log_3(x)\\ \quad \\
\cfrac{1}{3}=log_3(x)\implies \large 3^{\frac{1}{3}}=x\)

Answer 7

\(\large\bf 3^{\frac{1}{3}}=x\implies \sqrt[3]{3}=x\)

Answer 8

where do I go from here now? \[Log_3(1/3)=x?\]

Answer 9

I'm kind of lost on how we get to the end 3(1/3) part

Answer 10

what part?

Answer 11

After we get 1/3

Answer 12

\(\large \bf {\color{red}{\cfrac{1}{3}}=log_\color{blue}{3}(x)\implies \color{blue}{3}^{\color{red}{\frac{1}{3}}}=x\implies \sqrt[3]{3}=x}\)

Answer 13

all I did was convert it to exponent notation twice
once for the outer function
once for the inner one

Answer 14

so its 3^1/3?

Answer 15

yeap

Answer 16

and can be written at 3 root 3

Answer 17

yes

Answer 18

gotcha so one more question if you don't mind me asking

Answer 19

ok

Answer 20

Solve for X.

4 ^ (x- 4 ) = 3

Find an exact solution without decimals. You might want to use ln(x) or log(x).

Answer 21

\(\bf 4^{x- 4} = 3\\ \quad \\
\textit{using log cancellation rule of }\quad \color{blue}{log_aa^x=x}\qquad thus\\ \quad \\
4^{x- 4} = 3\implies log_4(4^{x- 4} )=log_4(3)\implies x-4=log_4(3)\\ \quad \\
\textit{using the log change of base rule of }\quad \color{blue}{log_ab=\cfrac{log_cb}{log_ca}}\qquad thus\\ \quad \\
x-4=log_4(3)\implies x-4=\cfrac{log_{10}3}{log_{10}4}\implies x=\cfrac{log_{10}3}{log_{10}4}+4\)

Answer 22

how did we get the log_4(4^x-4)

Answer 23

we take \(\bf log_4\) to both sides

Answer 24

the idea being to use the log cancellation rule
where the log base, MATCHES the base with the exponent

Answer 25

hence why they are both log base 10

Answer 26

why did I use the log change of base rule? well, technically I didn't have to
but since calculators don't have \(\bf log_3 \ or \ log_4\ only \ log_{10} \ or \ ln\), thus

Answer 27

so I typed in the answer and said I got it wrong :((

Answer 28

now you know why the suggestion was "You might want to use ln(x) or log(x)"

Answer 29

I typed log3/log4+4

Answer 30

that's good, I used \(\bf log_{10}\) just to make it non-ambiguous, but yes, log with no base implies base 10

Answer 31

ohhh hmmm, well...
I assume the answer expected is a value, so you have to get the logs first and add 4 to that

Answer 32

so just divide and add?

Answer 33

sorry for my lag :(

yes, you have to divide and add