How Do You Solve Exponential Equations with Different Bases?

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SUMMARY

The discussion focuses on solving the exponential equation (1/25)^(x+3) = 125^(2x). Participants clarify that both sides can be expressed with a common base of 5, specifically as 5^(-2(x+3)) = 5^(3(2x)). The correct approach involves equating the exponents after rewriting the bases, leading to a solvable equation for x. The importance of proper notation and parentheses in mathematical expressions is emphasized to avoid confusion.

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Homework Statement


(1/25)^x+3=125^x+3

Homework Equations


I have used log on both sides but keep getting an incorrect answer.

The Attempt at a Solution


How would solve this type of equation. I set the bases both to 5, make them equal to each other and i can't get the right answer.
 
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Is the + 3 in the exponent? If not you can substract it from both sides.
 
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If the exponent DOES include the + 3 , then the whole exponent should be enclosed in parentheses.

We follow standard Order of Operations here at PF .
 
yes the +3 is in the exponent. So i should put the whole thing in parentheses.
 
Make the substitution k = x + 3, then take the log of both sides.
 
Math_QED said:
Make the substitution k = x + 3, then take the log of both sides.
what is k. Is that the (1/25)?
 
So the equation is:

(1/25)^(x+3) = 125^(2x) ?

EDIT: The equation you initially posted seems to be not the same as the one you wrote down here. Never mind what I said about substitution.
 
Math_QED said:
So the equation is:

(1/25)^(x+3) = 125^(2x) ?

EDIT: The equation you initially posted seems to be not the same as the one you wrote down here. Never mind what I said about substitution.
yes that is the equation my apologies.
 
  • #10
johnsonjohn said:

Homework Statement


(1/25)^x+3=125^x+3

A company is valued at 5 million when first starting, one year later it is at 25 million. Write an exponential equation and find when it will reach 200 million value.

Homework Equations


I have used log on both sides but keep getting an incorrect answer.

The Attempt at a Solution


How would solve this type of equation. I set the bases both to 5, make them equal to each other and i can't get the right answer.

johnsonjohn said:
here a link to it.

Https://imgur.com/oPPO9zT
What you wrote in the first post is different from the image in your later post.
The equation apparently is ##(\frac 1 {25})^{x + 3} = 125^{2x}##

Link to our LaTeX primer: https://www.physicsforums.com/help/latexhelp/
 
  • #11
the equation (1/25)^(x+3)= 125^2X that's the equation on the paper.
 
  • #12
Mark44 said:
What you wrote in the first post is different from the image in your later post.
The equation apparently is ##(\frac 1 {25})^{x + 3} = 125^{2x}##

Link to our LaTeX primer: https://www.physicsforums.com/help/latexhelp/
sorry i didnt click on the link before i replied i will use the Latex thing next time.
 
  • #13
johnsonjohn said:
sorry i didnt click on the link before i replied i will use the Latex thing next time.

Regarding your equation. You must have seen that 1/25 and 125 can be written as 5^(-2) and 5^3
Rewrite your equation with this. Post what you get then.
 
  • #14
johnsonjohn said:
the equation (1/25)^(x+3)= 125^2X that's the equation on the paper.
If you write expression in that "in-line" sort of format, please use adequate parentheses.

For example:
(1/25)^(x+3)= 125^(2x)​

.
 
  • #15
I make it 5^(-2)(X+3)=5^(3)(2x) and this is where I think I mess up. I multiply the numbers in parentheses
 
  • #16
johnsonjohn said:
I make it 5^(-2)(X+3)=5^(3)(2x) and this is where I think I mess up. I multiply the numbers in parentheses

Now use: (a^b)^c = a^(bc)
 
  • #17
johnsonjohn said:
I make it 5^((-2)(X+3))=5^((3)(2x)) and this is where I think I mess up. I multiply the numbers in parentheses
f(x) = 5x is a one-to-one function.

Therefore, just equate the arguments of the two sides.
 
  • #18
Math_QED said:
Now use: (a^b)^c = a^(bc)
Sorry for starting up this thread again, but what do I put in that equation. Like is 5=A, I don't understand what numbers to plug in.
 
  • #19

johnsonjohn said:
Sorry for starting up this thread again, but what do I put in that equation. Like is 5=A, I don't understand what numbers to plug in.

You can ignore that. Once you have the expression in #16, you can say make an equation where you say the exponents are equal. (Do in fact you take the 5 bases log of what's left and right from the equality sign. Once you do that, you get an easy equation in x to solve. Let us know when you found the answer :)
 
  • #20
johnsonjohn said:
I make it 5^(-2)(X+3)=5^(3)(2x) and this is where I think I mess up. I multiply the numbers in parentheses
Yes, but incorporating the extra necessary parentheses as SammyS corrected for you in post #18.

➫ Please apply more care in your future threads. Careless mistakes and omissions in presentation don't give confidence that you are valuing the assistance of forum helpers.
 

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