Exponents with different bases

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Homework Help Overview

The discussion revolves around solving the equation 4^(x+2) = 9, which involves exponents with different bases. Participants are exploring methods to approach this type of problem, particularly focusing on the use of logarithms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of logarithms to solve the equation, with some expressing uncertainty about handling addition and subtraction in exponents. There are mentions of the Change-of-Base formula and its relevance to the problem. Questions arise regarding the correctness of specific logarithmic manipulations.

Discussion Status

The discussion is active, with various participants offering insights and questioning each other's reasoning. Some guidance on using logarithms has been provided, but there is no consensus on the correctness of all statements made, particularly regarding the manipulation of logarithmic expressions.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the methods they can use or the depth of their explorations. There is an emphasis on understanding the properties of logarithms in relation to the problem at hand.

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Homework Statement


4^(x+2)=9


Homework Equations


basically all I want to know is what is the best way to go about solving a problem like this.


The Attempt at a Solution


I know how to convert a problem as such 4^(3x)=10 using. [log(<base>4,10)]/3 = X. however I need some help with addition and subtraction in exponents with one base equaling another number with a different base.
 
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I don't see what the problem is. I mean on the LHS you have factors of 2 and on the RHS you have factors of 3, so you really can't simplify further in a sense.

The best way to go is using logarithms. It should be clear you won't find a "nice" value for x (obviously no integer solutions).
 
H0T_S0UP said:

Homework Statement


4^(x+2)=9


Homework Equations


basically all I want to know is what is the best way to go about solving a problem like this.


The Attempt at a Solution


I know how to convert a problem as such 4^(3x)=10 using. [log(<base>4,10)]/3 = X. however I need some help with addition and subtraction in exponents with one base equaling another number with a different base.

Well, as for the addition and subtraction, it seems you have the right idea by dividing that logarithm of base 4 by 3. It's the same idea, algebra is algebra. You can look at the log as a variable to help get your mind around it. (x+3 instead of log(<base>3,6)+3) And as for the different bases, there is a handy formula called the Change-of-Base formula! It will be your best friend with logarithms. Here is a quick copy and paste of it, you can find more if you search it using yahoo or google.

[tex]log_a(x)[/tex] can be converted to base 'b' by the formula
[tex](log_b(x))/(log_b(a))[/tex]

(sorry, not too familiar with LaTeX, but I hope you get the idea. As I said, you can just search for the formula. Best of luck!)
 
H0T_S0UP said:

Homework Statement


4^(x+2)=9


Homework Equations


basically all I want to know is what is the best way to go about solving a problem like this.


The Attempt at a Solution


I know how to convert a problem as such 4^(3x)=10 using. [log(<base>4,10)]/3 = X. however I need some help with addition and subtraction in exponents with one base equaling another number with a different base.

It doesn't matter what base you use: if 4x+2= 9 then log(4x+2)= (x+2)log 4= log 9 where "log" can be to any base. I presume your calculator has both base 10 and natural logarithms so use those and solve for x.
 
Using logs I've found that X=(log9/log4)-2 = log9-log4-2. Is this correct?
 
No! your final statement

[tex] \frac{\log 9}{\log 4} - 2 = \log 9 - \log 4 - 2[/tex]

is not correct.

It is true that

[tex] \log \left( \frac 9 4 \right) = \log 9 - \log 4[/tex]

but that is not what your answer involves.
 

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