# Logarithm

• Sep 17th 2009, 04:50 AM
dzein
Logarithm
Hello, can you please slove the logarithm? I don't understand how to open it.

$\displaystyle Log_7[x+log_2(9-2^x)+4]=1$

I tried this
$\displaystyle Log_7(x+4)+Log_9(9-2^x)=1$
But not sure if it is right and don't know what to do next.

EDIT:Oops, wrong topic :x
• Sep 17th 2009, 05:01 AM
stapel
Quote:

Originally Posted by dzein
I tried this
$\displaystyle Log_7(x+4)+Log_9(9-2^x)=1$
But not sure if it is right....

Um... no, it's not right. (Surprised)

Have you studied logarithms at all? (The fact that you think bases can be added suggests not, but I don't want to assume.)
• Sep 17th 2009, 05:03 AM
dzein
Nope, I missed some classes due to sickness :/
• Sep 17th 2009, 05:12 AM
stapel
You note that you have not studied logs yet. (I guess you were sick for a couple of weeks?)

The step-by-step instructions won't make any sense until you learn the material you missed, so that's where you'll need to start.

Logarithms
Log Rules
Solving Log Equations
Solving Exponential Equations

After you've studied the above, you should be ready to at least get started on this exercise:

Use "The Relationship" to convert the log-base-seven equation to the equivalent exponential form. Subtract the 4 and the x over to the right-hand side. Simplify.

Use "The Relationship" again to covert the log-base-two equation to the equivalent exponential form.

• Sep 17th 2009, 05:42 AM
mr fantastic
Quote:

Originally Posted by dzein
Hello, can you please slove the logarithm? I don't understand how to open it.

$\displaystyle Log_7[x+log_2(9-2^x)+4]=1$

I tried this
$\displaystyle Log_7(x+4)+Log_9(9-2^x)=1$
But not sure if it is right and don't know what to do next.

EDIT:Oops, wrong topic :x

$\displaystyle \log_7 [x + \log_2 (9-2^x) + 4] = 1$

$\displaystyle \Rightarrow x + \log_2 (9-2^x) + 4 = 7$

$\displaystyle \Rightarrow \log_2 (9-2^x) = 3 - x$

By inspection, x = 0 and x = 3 are the solutions.
• Sep 17th 2009, 12:24 PM
Krizalid
Quote:

Originally Posted by mr fantastic

$\displaystyle \Rightarrow \log_2 (9-2^x) = 3 - x$

Following up on this, we have $\displaystyle 9-2^{x}=\frac{8}{2^{x}},$ so $\displaystyle 2^{2x}-9\cdot 2^{x}+8=\left( 2^{x}-1 \right)\left( 2^{x}-8 \right),$ and we should be able to finish this from here. (This agrees with mr fantastic's solutions.)
• Sep 17th 2009, 10:23 PM
pacman
if f(x) = 2^(3 - x) + 2^x - 9, then the plot agrees with the roots obtained above, x = 0, 3