# Homework Help: Need help with Logs

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1. Apr 22, 2016

### johnsonjohn

1. The problem statement, all variables and given/known data
(1/25)^x+3=125^x+3

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

3. 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 cant get the right answer.

Last edited by a moderator: Apr 30, 2016
2. Apr 22, 2016

### Math_QED

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

Last edited by a moderator: Apr 30, 2016
3. Apr 22, 2016

### SammyS

Staff Emeritus
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. Apr 22, 2016

### johnsonjohn

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

5. Apr 22, 2016

### Math_QED

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

6. Apr 22, 2016

### johnsonjohn

7. Apr 22, 2016

### johnsonjohn

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

8. Apr 22, 2016

### Math_QED

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. Apr 22, 2016

### johnsonjohn

yes that is the equation my apologies.

10. Apr 22, 2016

### Staff: Mentor

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. Apr 22, 2016

### johnsonjohn

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

12. Apr 22, 2016

### johnsonjohn

sorry i didnt click on the link before i replied i will use the Latex thing next time.

13. Apr 22, 2016

### Math_QED

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. Apr 22, 2016

### SammyS

Staff Emeritus
If you write expression in that "in-line" sort of format, please use adequate parentheses.

For example:
(1/25)^(x+3)= 125^(2x)​

.

15. Apr 22, 2016

### 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

16. Apr 22, 2016

### Math_QED

Now use: $(a^b)^c$ = a^(bc)

17. Apr 22, 2016

### SammyS

Staff Emeritus
f(x) = 5x is a one-to-one function.

Therefore, just equate the arguments of the two sides.

18. Apr 29, 2016

### johnsonjohn

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. Apr 29, 2016

### Math_QED

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. Apr 30, 2016

### Staff: Mentor

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.