Logarithm

**dan123** · April 27th 2009, 12:24 PM

solve Log5(x-4)= Log7 X for X. Round your answer to four decimal places

and

for log y= log(0.5x-3)+ Log 2 state the domain,and express y as a function of x

anyone know?

**Jameson** · April 27th 2009, 12:57 PM

Originally Posted by dan123

solve Log5(x-4)= Log7 X for X. Round your answer to four decimal places

and

for log y= log(0.5x-3)+ Log 2 state the domain,and express y as a function of x

Hi dan123,

Welcome to MHF! Please post only one question per thread in the future.

I am assuming by "log" you mean $\log_{10}$ . In higher math, this can mean $\log_{e}=\ln$ so be careful and specific.

I'll help with your second question. Notice that log(x) has a real domain for x>0, not including 0. So use that to make some deductions. Firstly, 0.5x-3 > 0 for this to work but also log(0.5x-3) > 0 so that log(y) is defined. For what values of "a" is log(a) negative?

**dan123** · April 27th 2009, 01:07 PM

I figured out the second one,I just struggled with the domain,anything on the first one,I can't seem to figure it out for the life of me

**Jameson** · April 27th 2009, 01:11 PM

Originally Posted by dan123

I figured out the second one,I just struggled with the domain,anything on the first one,I can't seem to figure it out for the life of me

Really? Are you sure you have the problem solved? You can post your solution, including your steps here to be sure. It's not that simple of a question.

Originally Posted by dan123

solve Log5(x-4)= Log7 X for X. Round your answer to four decimal places

I would use this nice change of base formula:

$\log_{a}(b)=\frac{\ln(b)}{\ln(a)}$ .

Change both sides using this method then combine terms using log rules.

**dan123** · April 27th 2009, 01:30 PM

so in my history of the change of base...would I go....

log base 7 X

log base 10 x/ Log base 10 7

take out the 10s....

logx/log 7??

**stapel** · April 28th 2009, 06:19 AM

Originally Posted by dan123

log base 10 x/ Log base 10 7

take out the 10s....

I'm sorry, but what do you mean by "taking out the 10s"...? (You can't have a log without a base. Are you perhaps referring to restating the expression without the "understood" common-log base...?)

Thank you!

**dan123** · April 29th 2009, 01:10 PM

I don't exactly a hundred percent understand the change of base formula,could you show me an example question?

would I go log base 5(x-4)/log base 7 x

**e^(i*pi)** · April 29th 2009, 01:37 PM

Originally Posted by dan123

I don't exactly a hundred percent understand the change of base formula,could you show me an example question?

would I go log base 5(x-4)/log base 7 x

Using the change of base rule:

Simple example: $log_2(5) = \frac{log_{10}(5)}{log_{10}(2)}$

More complicated example: $log_9(36) = \frac{ln(36)}{ln(9)} = \frac{ln(4)ln(9)}{ln(9)} = ln(4) = ln(2^2) = 2ln(2)$

You may choose to change to any base which is greater than 1 but most commonly you'd switch to either base 10 (first example) or base e (second example) as all calculators should have both these on

**dan123** · April 29th 2009, 04:41 PM

so changing the bases should give you a number,and use that number to solve the equation

**e^(i*pi)** · April 29th 2009, 05:08 PM

Originally Posted by dan123

so changing the bases should give you a number,and use that number to solve the equation

The change of base rule will allow you to pick a base supported by your calculator - usually base 10 or base e.

log base 7 X = $log_7(X)$

As is your calculator probably doesn't have a base 7 logarithm button (if it does feel free to evaluate it directly unless you're told to use the change of base)

The problem above can be converted to base 10, which should be on your calculator so you can punch in the numbers and solve:

$log_7(X) = \frac{log_{10}(X)}{log_{10}(7)}$

**dan123** · April 29th 2009, 05:39 PM

so essentially that is the solution to the problem,change the base to a 10,divide them,and the number is the solution

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