How Do You Solve Exponential Equations with Different Bases?

[johnsonjohn]

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.

 

[member 587159]

Is the + 3 in the exponent? If not you can substract it from both sides.

 

[SammyS]

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 .

 

[johnsonjohn]

yes the +3 is in the exponent. So i should put the whole thing in parentheses.

 

[member 587159]

Make the substitution k = x + 3, then take the log of both sides.

 

[johnsonjohn]

here a link to it.

Https://imgur.com/oPPO9zT

 

[johnsonjohn]

Make the substitution k = x + 3, then take the log of both sides.

what is k. Is that the (1/25)?

 

[member 587159]

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.

 

[johnsonjohn]

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.

 

[Mark44]

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.

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/

 

[johnsonjohn]

the equation (1/25)^(x+3)= 125^2X that's the equation on the paper.

 

[johnsonjohn]

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.

 

[member 587159]

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.

 

[SammyS]

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)​

.

 

[johnsonjohn]

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

 

[member 587159]

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)

 

[SammyS]

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) = 5^x is a one-to-one function.

Therefore, just equate the arguments of the two sides.

 

[johnsonjohn]

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.

 

[member 587159]

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 :)

 

[NascentOxygen]

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.

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