1. Some Logarithm Questions

If anyone could help me with these, that would be great.

3. Solve each equation
(a) 7^(x-1) = 49^(4-2x)
(b) log3x = 3/2log39 + log32
(c) log (x+1) + log (x-5) = 1

4. Solve each equation to three decimal places
(a) 4^x = 24
(b) 3^(2x) = 8
(c) 5^(x-2) = 150

5. On the same grid, sketch the graph of each function
(a) y = log3x
(b) y = 3^x
(c) y = 2(0.5)^x - 6

2. Hello, Philip!

I must assume you know the basic properities of logs . . .

2. Express as a single logarithm

$(a)\;\log(x) - 4\log(y)$

$(b)\;6\log_3(x) - \frac{1}{2}\log_3(y) + \log_3(2x)$

(a) We have: . $\log(x) - 4\log(y) \;=\;\log(x) - \log(y^4) \;=\;\log\left(\frac{x}{y^4}\right)$

(b) We have: . $6\log_3(x) - \frac{1}{2}\log_3(y) + \log_3(2x)\;=\;\log_3\left(x^6\right) - \log_3\left(y^{\frac{1}{2}}\right) + \log_3(2x)$
. . $= \;\log_3\left(\frac{x^6\cdot2x}{y^{\frac{1}{2}}}\r ight) \;=\;\log_3\left(\frac{2x^7}{y^{\frac{1}{2}}}\righ t)$

3. Solve each equation

$(a)\;7^{x-1} \:= \:49^{4-2x}$

$(b)\;\log_3(x) \:= \:\frac{3}{2}\log_3(9) + \log_3(2)$

$(c)\;\log(x+1) + \log(x-5) \:= \:1$

(a) We have: . $7^{x-1} \:=\:49^{4-2x}$

Get the same base on both sides: . $7^{x-1} \:=\:(7^2)^{4-2x} \:=\:7^{2(4-2x)}$

. . And we have: . $7^{x-1}\:=\:7^{8 - 4x}$

Since the bases are equal, then the exponents are equal.
. . $x - 1 \:=\:8 - 4x\quad\Rightarrow\quad x = \frac{9}{5}$

(b) We have: . $\log_3(x) \:=\:\frac{3}{2}\log_3(9) + \log_3(2) \:=\:\log\left(3^{\frac{3}{2}}\right) + \log_3(2)\:=\:\log_3(27) + \log_3(2)$

. . $\log_3(x) \:=\:\log_3(27\cdot2)\quad\Rightarrow\quad\log_3(x ) \:=\:\log_3(54)$

Since the bases are equal, the "arguments" are equal: . $x \,=\,54$

(c) We have: . $\log(x+1) + \log(x-5)\:=\:1\quad\Rightarrow\quad\log[(x+1)(x-5)] \:=\:1$

Then: . $\log(x^2 - 4x - 5) \:=\:1\quad\Rightarrow\quad x^2 - 4x - 5 \:=\:10^1$

. . We have a quadratic: . $x^2 - 4x - 15 \:=\:0$

. . Quadratic Formula: . $x \:=\:\frac{4 \pm\sqrt{76}}{2} \:=\:2 \pm\sqrt{19}$

We find that $x \:= \:2 - \sqrt{19}$ is an extraneous root.

Therefore, the only solution is: . $x \,=\,2 + \sqrt{19}$

3. Originally Posted by philipsach1
If anyone could help me with these, that would be great.

3. Solve each equation
(a) 7^(x-1) = 49^(4-2x)
(b) log3x = 3/2log39 + log32
(c) log (x+1) + log (x-5) = 1

4. Solve each equation to three decimal places
(a) 4^x = 24
(b) 3^(2x) = 8
(c) 5^(x-2) = 150

5. On the same grid, sketch the graph of each function
(a) y = log3x
(b) y = 3^x
(c) y = 2(0.5)^x - 6
Philip, please don't remove the questions that have been answered from your original post. If someone else has questions on the same topic then they can benefit from them. If you remove the questions it meakes reading the thread very confusing! Thanks.

-Dan

4. Originally Posted by philipsach1
4. Solve each equation to three decimal places
(a) 4^x = 24
(b) 3^(2x) = 8
(c) 5^(x-2) = 150
These are all essentialy the same take logs of both sides and solve the
resulting equation.

I will do (c) to show the principle:

log(5^(x-2))=log(150)

so using the property that log(a^b)=b log(a):

(x-2)log(5)=log(150),

x-2=log(150)/log(5)

x=log(150)/log(5)+2

Now getting out a calculator we have (note it does not matter what
base you take for the logs you will get the same answer, but I used
base e, it will work just as well with base 10):

x~= 5.113

RonL

5. Originally Posted by philipsach1
4. Solve each equation to three decimal places
(a) 4^x = 24
(b) 3^(2x) = 8
(c) 5^(x-2) = 150
a) $4^x = 24$

$log_4(4^x) = log_4(24)$

$x = log_4(24)$

Unless you have a really fancy calculator you can't calculate this directly. I'm going to use the "change of base" formula to change the base from 4 to 10:
$log_b(a) = \frac{log_c(a)}{log_c{b}}$ where c is your new base.

So $log_4(24) = \frac{log(24)}{log(4)} \approx 2.29248$, where "log" is $log_{10}$.

If your calcuator doesn't have a "log" key it will almost certainly have a "ln" key, the natural log, or log base e. You can freely use that one instead if you like.

b) $3^{(2x)} = 8$

A different way. We don't need to use $log_3$ here, we can use a base that we can use directly on the calculator. In this example I'll use the "ln" function:
$ln(3^{2x}) = ln(8)$

By the properties of logarithms:
$2x \cdot ln(3) = ln(8)$

$x = \frac{ln(8)}{2 \cdot ln(3)} \approx 0.946395$
(You can verify this is the same answer we would get if you used $log_3$ and then used the change of base formula.)

c) $5^{(x-2)} = 150$
(Note, I'm a Physicist, so I tend to use ln a lot.)
$ln(5^{x-2} = ln(150)$

$(x - 2) \cdot ln(5) = ln(150)$

$x = 2 + \frac{ln(150)}{ln(5)} \approx 5.11328$

-Dan

6. thanks

thanks to everyone who has helped so far.

could someone pleez show me what the graphs look like for question number 5 so that i know mine are right?

7. Originally Posted by philipsach1
thanks to everyone who has helped so far.

could someone pleez show me what the graphs look like for question number 5 so that i know mine are right?
Just for clarity, the red graph is the $log_3(x)$.

-Dan

8. Originally Posted by philipsach1
...

7. Express the function y = 5^x as an equivalent function with base 3.

...
Hello,

a very important formula with logarithm is:

$x=b^{\log_{b}{(x)}}$. Therefore:

$5=3^{\log_{3}{(5)}}$

There is only one difficulty: Your calculator can't calculate directly logarithms to the base 3. Therefore use this formula (log means base 10):

$\log_{3}{(x)}=\frac{\log(x)}{\log(3)}$

$y=3^{\frac{\log(5)}{\log(3)} \cdot x}$

EB

9. Originally Posted by philipsach1
I'm sorry, these questions are posted elsewhere, but I can't really get any detailed answers.

7. Express the function y = 5^x as an equivalent function with base 3.

9. The decibel (dB) is used to measure the loudness of a sound. The equation D = 10 log I models the decibel level of a sound whose intensity is I watts per square metre (W/m^2). The decibel levels of a subway train and normal conversation are 115 dB and 60 dB, respectively. How many times as intense as normal conversation is the noise of a subway train?
Both of these have been aswered to the best of the helpers ability elsewhere.

If you dont understand ask for more help in those threads and don't
repost the questions. You are wasting our time.

RonL