Logarithm problems

[Faith54]

I have a few log problems that I'm stuck on. Any help would be appreciated.

(log 3 x)squared - log 3 xsquared = 3

(log a x) to the -1 = log a x to the -1

log 7 square root of (x squared - 9) = 1

x log 1/6 = log 6 (The base is 10 for this one, not 6)

I know it's kind of hard to read since I don't know how to copy the problems directly, so if you don't understand what I'm trying to say let me know.

 

[galactus]

I have a few log problems that I'm stuck on. Any help would be appreciated.

(log 3 x)squared - log 3 xsquared = 3

(log a x) to the -1 = log a x to the -1

log 7 square root of (x squared - 9) = 1

I assume you mean base 7. What this means is

\(\displaystyle log_{7}\sqrt{x^{2}-9}=1\rightarrow\7^{1}=\sqrt{x^{2}-9}\)

Solve for x

x log 1/6 = log 6 (The base is 10 for this one, not 6)

I know it's kind of hard to read since I don't know how to copy the problems directly, so if you don't understand what I'm trying to say let me know.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\left( {\log _3 (x)} \right)^2 - \log _3 (x^2 ) = 3 \\
\left( {\log _3 (x)} \right)^2 - 2\log _3 (x) - 3 = 0 \\
u = \log _3 (x)\quad \Rightarrow u^2 - 2u - 3 = 0 \\
(u - 3)(u + 1) = 0\quad \Rightarrow \quad u = 3 \vee u = - 1 \\
\log _3 (x) = 3\quad \Rightarrow \quad x = ? \\
\end{array}\)

 

[soroban]

Hello, Faith54 !

\(\displaystyle 1)\;\;[\log_3(x)]^2\,-\,\log_3(x^2)\;=\;3\)

We have: \(\displaystyle \:[\log_3(x)]^2\,-\,2\cdot\log_3(x)\,-\,3\;=\;0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \:[\log_3(x)\,+\,1]\,[\log_3(x)\,-\,3]\;=\;0\)


Then: \(\displaystyle \:\log_3(x)\,+\,1\:=\:0\;\;\Rightarrow\;\;\log_3(x)\:=\:-1\;\;\Rightarrow\;\;x\,=\,3^{-1}\,=\,\frac{1}{3}\)

And: \(\displaystyle \:\log_3(x)\,-\,3\:=\:0\;\;\Rightarrow\;\;\log_3(x)\,=\,3\;\;\Rightarrow\;\;x\,=\,3^3\,=\,27\)

Both answers check out!

\(\displaystyle 2)\;\;[\log_a(x)]^{-1}\;=\;\log_a(x^{-1})\)

We have: \(\displaystyle \L\,\frac{1}{\log_a(x)}\)\(\displaystyle \;=\;-1\cdot\log_a(x)\;\;\Rightarrow\;\;1\;=\;-[\log_a(x)]^2\)

Then: \(\displaystyle \:[\log_a(x)]^2\:=\:-1\;\;\Rightarrow\;\;\log_a(x)\:=\:\sqrt{-1}\;\) ?? . . . No real solution

\(\displaystyle 3)\;\;\log_7\sqrt{x^2\,-\,9}\;=\;1\)

Then: \(\displaystyle \:\sqrt{x^2\,-\,9}\:=\:7^1\)

Square both sides: \(\displaystyle \,x^2\,-\,9\:=\:49\;\;\Rightarrow\;\;x^2\,=\,58\)

Therefore: \(\displaystyle \,x\,=\,\pm\sqrt{58}\)

\(\displaystyle 4)\;\;x\cdot\log\left(\frac{1}{6}\right)\:=\;\log(6)\) . . . (base 10)

We have: \(\displaystyle \:x\cdot\log(6^{-1})\:=\:\log(6)\)

Then we have: \(\displaystyle \:x\cdot(-1)\cdot\log(6)\:=\:\log(6)\)

Divide by -\(\displaystyle \log(6):\;\;x\:=\:-1\)

 

[phrisma]

Suppose that a student has 500 vocabulary words to learn. If the student learns 15 words after 5 minutes (this means L(5)=15 for the function L(t) ) , the function
L(t) = 500(1− e−0.0061t ) approximates the number of words L that a student will learn after t minutes.
a) How many words will the student learn after 30 minutes?
b) How long will it take the student to learn 300 words?

If I just replace the values in the equation the answer is negative for the first question, that is correct?
Could you please, help me.

 

[happy]

Suppose that a student has 500 vocabulary words to learn. If the student learns 15 words after 5 minutes (this means L(5)=15 for the function L(t) ) , the function
L(t) = 500(1− e−0.0061t ) approximates the number of words L that a student will learn after t minutes.
a) How many words will the student learn after 30 minutes?
b) How long will it take the student to learn 300 words?

If I just replace the values in the equation the answer is negative for the first question, that is correct?
Could you please, help me.

Start your own thread, would you?

 

[phrisma]

L(t) = 500 (1 - e ^ -o.0061 t)
L(30) = 500 (1 - e ^ 0.0061*30)
L(30) = 500 (1 - e ^ -1.83)

L(30) = 500 (1 - 0.1604) is this correct?
L(30) = 500 (0.8396)
L(30) = 419.8[/tex][/url][/code]

 

[galactus]

\(\displaystyle \L\\300=500(1-e^{-.0061t})\)

Solve for t:

\(\displaystyle \L\\\frac{3}{5}=1-e^{-.0061t}\)

\(\displaystyle \L\\\frac{2}{5}=e^{-.0061t}\)

\(\displaystyle \L\\ln(\frac{2}{5})=-.0061t\)

\(\displaystyle \L\\\frac{ln(\frac{2}{5})}{-.0061}=\frac{10000ln(\frac{5}{2})}

{61}\)=150.21159......