Exponents with different bases

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To solve the equation 4^(x+2)=9, logarithms are the most effective method, as it is unlikely to yield a simple integer solution. The Change-of-Base formula is essential for working with different bases in logarithmic equations. By applying logarithms, the equation can be transformed into log(4^(x+2)) = log(9), leading to (x+2)log(4) = log(9). It's important to note that the final expression should be simplified correctly, avoiding mistakes in combining logarithmic terms. Using a calculator with logarithmic functions will aid in finding the value of x accurately.
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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.

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

(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

<br /> \frac{\log 9}{\log 4} - 2 = \log 9 - \log 4 - 2<br />

is not correct.

It is true that

<br /> \log \left( \frac 9 4 \right) = \log 9 - \log 4<br />

but that is not what your answer involves.
 
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