Logarithm

**dzein** · September 17th 2009, 04:50 AM

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

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

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

EDIT:Oops, wrong topic :x

**stapel** · September 17th 2009, 05:01 AM

Originally Posted by dzein

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

Um... no, it's not right.

Have you studied logarithms at all? (The fact that you think bases can be added suggests not, but I don't want to assume.)

**dzein** · September 17th 2009, 05:03 AM

Nope, I missed some classes due to sickness :/

**stapel** · September 17th 2009, 05:12 AM

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.

Then you'll need to get fancy. If you get stuck at that stage, please reply showing all of your steps so far, and we'll be glad to help you with the quadratic-form base-two exponential equation.

**mr fantastic** · September 17th 2009, 05:42 AM

Originally Posted by dzein

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

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

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

EDIT:Oops, wrong topic :x

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

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

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

By inspection, x = 0 and x = 3 are the solutions.

**Krizalid** · September 17th 2009, 12:24 PM

Originally Posted by mr fantastic

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

Following up on this, we have $9-2^{x}=\frac{8}{2^{x}},$ so $2^{2x}-9\cdot 2^{x}+8=\left( 2^{x}-1 \right)\left( 2^{x}-8 \right),$ and we should be able to finish this from here. (This agrees with mr fantastic's solutions.)

**pacman** · September 17th 2009, 10:23 PM

if f(x) = 2^(3 - x) + 2^x - 9, then the plot agrees with the roots obtained above, x = 0, 3

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