Solving for x in an Exponential Equation

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Homework Help Overview

The discussion revolves around solving for x in the exponential equation (3.4)^(2x+3) = 8.5. Participants are exploring different interpretations of the equation and the methods to solve it.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to clarify the original equation's format, with suggestions that it could be interpreted in multiple ways. Some are exploring the use of logarithms to solve the equation, while others are questioning the steps taken to manipulate the expressions.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided guidance on using logarithms, while others are questioning the correctness of the transformations applied to the equations. There is no clear consensus on the correct approach or solution at this point.

Contextual Notes

There is uncertainty regarding the original equation's structure, which affects the subsequent calculations and interpretations. Participants are also emphasizing the importance of using parentheses for clarity in mathematical expressions.

steveandy2002
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I need to find x from (3.4)^2x+3=8.5
what i have done is as follows
3.4^2x = 11.56
3.4^2x+3 = 14.56
8.5 - 14.56 = x
x=-6.06 <<<<is this the correct way of solving and correct answer??
 
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Hi SteveAndy
Your formula is a bit confusing:

(3.4)^2x+3=8.5

Should the formula look like this:

1. 3.4(2x+3)=8.5

or

2. 3.42x+3=8.51 or 2?
 
Last edited:
if it is :

3.4(2x+3)=8.5


take the log of both sides...
 
There is even a third possibility:
3.42x + 3 = 8.5

steveandy2002 said:
I need to find x from (3.4)^2x+3=8.5
what i have done is as follows
3.4^2x = 11.56
3.4^2x+3 = 14.56
8.5 - 14.56 = x
x=-6.06
Putting aside the uncertainty of what the original equation actually is, how did you go from the first equation to the second? The 11.56 in the second equation happens to be 3.42, but the second equation should not involve x.
 
huntoon said:
if it is :

3.4(2x+3)=8.5


take the log of both sides...

so based on useing logs would the folllowing be correct??
3.4^(2x+3)=8.5
2x+3log3.4=log8.5
2x=3-log8.5/log3.4
x= 3-log8.5/log3.4/2
x=1.947942814?
 
steveandy2002 said:
so based on useing logs would the folllowing be correct??
3.4^(2x+3)=8.5
2x+3log3.4=log8.5
2x=3-log8.5/log3.4
x= 3-log8.5/log3.4/2
x=1.947942814?

Should be:
3.4(2x+3)=8.5

exponent rule first:
3.4(2x+3) becomes 3.42x * 3.43

so...
3.42x * 3.43 = 8.5

so...
3.42x = 8.5/3.43

Now log both sides...
ln(3.42x) = ln(8.5/3.43)

apply log rule:
2xln(3.4) = ln(8.5/3.43)

therefore:
x = ln(8.5/3.43)/(2ln(3.4))

I get approx: -0.63 for x.


and to check:

3.4(2(-0.63)+3)=8.5
3.4(1.74)=8.5
8.4 = 8.5

hmm... too much rounding, but close enough!
 
PLEASE USE PARENTHESES!
steveandy2002 said:
so based on useing logs would the folllowing be correct??
3.4^(2x+3)=8.5
2x+3log3.4=log8.5
The equation above is incorrect. It should be
(2x + 3)log3.4 = log8.5
steveandy2002 said:
2x=3-log8.5/log3.4
This is wrong as well.
Starting from (2x + 3)log3.4 = log8.5 there is no way you can get to 2x=3-log8.5/log3.4
steveandy2002 said:
x= 3-log8.5/log3.4/2
x=1.947942814?
 

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