# Thread: Log and Exponent Questions

1. ## Log and Exponent Questions

Here are five log and exponent questions. I'm wasn't sure where to post them at so if this isn't the right place to make this thread, could someone please move it to where its proper place is.
Anyways, I've been able to solve some of these but I got decimals which is bad and I'm sure that I'm doing something wrong. I understand logs, these questions are just tricky because they involve more factoring and square roots.
Here's a picture:

If that doesn't work then here's a link:
http://img329.imageshack.us/img329/1750/zexp59ri8.png

2. Originally Posted by SportfreundeKeaneKent
Here are five log and exponent questions. ...
Hello,

a product of two factors equals zero if one of the factors equals zero. Thus

$5^{2x}-7=0\ \vee \ 5^x+6=0$

5^x > 0 for all x in IR. Thus there isn't a real solution for the 2nd equation.

$5^{2x}-7=0 \Longleftrightarrow 5^{2x}=7 \Longleftrightarrow 2x=\log_5{7} \Longleftrightarrow x=\frac{1}{2} \log_5{7}$

To calculate this value:

$x=\frac{1}{2} \log_5{7}=\frac{\ln(\sqrt{7})}{\ln5} \approx 0.60453...$

3. Originally Posted by SportfreundeKeaneKent
Here are five log and exponent questions. ...
Hello,

use the log-properties to simplify this equation:

$\log_3(w-6) + \log_3(w-14)=2, w > 14$

$\log_3\left( (w-6) (w-14) \right)=2$

$(w-6) (w-14)=3^2$. This is a quadratic equation in w. Expand the LHS:

$w^2-20w+75=0$. Use the formula to solve this equation. You'll get: w = 15 or w = 5. 5 is not defined for this equation. Thus the only possible solution is w = 15.

EB

4. Originally Posted by SportfreundeKeaneKent
Here are five log and exponent questions. ...
Hello Sportsfreund,

$9^x-9^{x-1}=648$. Use the properties of powers: $9^x-\frac{1}{9} \cdot 9^{x}=648$. Factorize:
$9^x \cdot \left( 1-\frac{1}{9} \right)=648$. Divide:
$9^x =729 \Longleftrightarrow 9^x=9^3$. Thus x = 3.

Plug in this value into the given term:

$(2 \cdot 3)^3=216$

EB

5. Originally Posted by SportfreundeKeaneKent
Here are five log and exponent questions. ...
Hello Sportsfreund,

$10000=5000 \cdot (1.05)^n \Longleftrightarrow 2=(1.05)^n$. Use logarithms:

$\ln(2)=n \cdot \ln(1.05) \Longleftrightarrow n=\frac{\ln(2)}{\ln(1.05)}\approx 14.206699... \approx 14.21$

EB

6. Originally Posted by SportfreundeKeaneKent
Here are five log and exponent questions. ...
Hello Sportsfreund,

$\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\fra c{4^{10}(2^{10}+1)}{4^4(2^4+4^7)}}
= \sqrt{4^6 \cdot \frac{(1025)}{(16400)}}=\sqrt{4^6 \cdot \frac{1}{16}}=16$

EB

7. Lovely work, EB!

In the last one, I changed everything to powers-of-2.
. . (It seemed to be the moral thing to do.)

$\sqrt{\frac{8^{10} + 4^{10}}{8^4 + 4^{11}}} \;=\;\sqrt{\frac{(2^3)^{10} + (2^2)^{10}}{(2^3)^4 + (2^2)^{11}}} \;=\;\sqrt{\frac{2^{30} + 2^{20}}{2^{12} + 2^{22}}}$

. . $= \;\sqrt{\frac{2^{20}(2^{10} + 1)}{2^{12}(2^{10} + 1)}} \;=\;\sqrt{\frac{2^{20}}{2^{12}}} \;=\;\sqrt{2^8} \;=\;2^4 \;=\;16

$

8. The second question is factorable to:
(w-15)(w-5)=0
So the answers are 15 and 5 are both answers, not just 15.

9. Originally Posted by SportfreundeKeaneKent
The second question is factorable to:
(w-15)(w-5)=0
So the answers are 15 and 5 are both answers, not just 15.
Hello Sportsfreund,

of course you are right that the solutions of this quadratic equation are w = 15 or w = 5. But as I've mentioned above the "solution" w = 5 doesn't satisfy your original equation.

Plug in 5:

$\log_3(w-6)+\log_3(w-14)=2$ will become:

$\log_3(5-6)+\log_3(5-14)=2$

$\log_3(-1)+\log_3(-9)=2$. But you can't calculate the logarithms of negative numbers; thus w = 5 is not a solution of your equation.

EB