Solving for x in an Exponential Equation

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The discussion revolves around solving the exponential equation (3.4)^(2x+3) = 8.5. Participants clarify the correct interpretation of the equation, with suggestions to use logarithms for solving. One user derives an approximate solution of x = -0.63 after applying logarithmic rules, while another user questions the validity of the steps taken to reach x = 1.947942814. There is a consensus on the importance of using parentheses for clarity in mathematical expressions. The conversation emphasizes the need for precise notation and correct application of logarithmic properties in solving exponential equations.
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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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