How Do You Solve Exponential Equations with Different Bases?

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In summary: They are here to help you of their own free time, not to do the work that you should be doing.You're welcome. I think you should thank SammyS for helping you complete this thread. If you need help, then ask for it, but don't be lazy and make others do the work for you. It's your homework, not ours. If you have a question, ask it after you have done your best to solve the problem yourself. Don't expect others to do your work for you. In the end, it will be you who reaps the rewards for your work. If you don't do the work, you won't reap the rewards.
  • #1
johnsonjohn
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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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  • #2
Is the + 3 in the exponent? If not you can substract it from both sides.
 
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  • #3
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 .
 
  • #4
yes the +3 is in the exponent. So i should put the whole thing in parentheses.
 
  • #5
Make the substitution k = x + 3, then take the log of both sides.
 
  • #7
Math_QED said:
Make the substitution k = x + 3, then take the log of both sides.
what is k. Is that the (1/25)?
 
  • #8
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.
 
  • #9
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 [itex]5^3[/itex]
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: [itex](a^b)^c [/itex] = 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: [itex](a^b)^c [/itex] = 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.
 

FAQ: How Do You Solve Exponential Equations with Different Bases?

What is the value of x in the equation (1/25)^x+3=125^x+3?

The value of x in this equation cannot be determined as it is not a valid mathematical expression. The equation should be written as (1/25)^x + 3 = (125)^x + 3 to accurately represent the use of exponents.

How can I solve the equation (1/25)^x+3=125^x+3?

This equation can be solved by taking the logarithm of both sides. The resulting equation will be x(log(1/25)) + 3 = x(log(125)) + 3. From here, you can isolate the x variable and solve for its value.

What is the domain of the equation (1/25)^x+3=125^x+3?

The domain of this equation is all real numbers, as there are no restrictions on the values of x that can be used in this equation.

How do I graph the equation (1/25)^x+3=125^x+3?

To graph this equation, you can use a graphing calculator or plot points by assigning values to x and solving for y. The resulting graph will be a straight line with a y-intercept of 3 and a slope of (log(125) - log(1/25)).

Is there a simpler way to write the equation (1/25)^x+3=125^x+3?

Yes, this equation can be simplified to (1/5)^x = 5^x by dividing both sides by (1/25)^3. This equation is equivalent to the original equation and can be solved in the same way.

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