Exponential Equation with Logarithms: Solving 4^x + 4^x+1 = 40

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Homework Help Overview

The discussion revolves around solving the exponential equation 4^x + 4^(x+1) = 40, with participants exploring various approaches to tackle the problem using logarithms and algebraic manipulation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use logarithms but expresses confusion about how to proceed. Some participants suggest rewriting the equation to factor out common terms, while others clarify the equation's structure. There are questions about the interpretation of the equation and the steps taken to simplify it.

Discussion Status

The discussion is active, with participants providing hints and clarifications. There is a mix of understanding and confusion, particularly regarding the steps taken to simplify the equation and the implications of those steps. Some guidance has been offered, but multiple interpretations of the problem are still being explored.

Contextual Notes

Participants are navigating the complexities of exponential equations and logarithmic properties, with some expressing uncertainty about the transformations applied to the original equation. The original poster's struggle with the problem reflects the challenges of working with exponential terms.

ms. confused
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Hi! I'm not sure how I would tackle this exponential equation:

4^x + 4^x+1 = 40

I was using logs to try and solve it but I'm getting nowhere. I don't know what to do exponentially either. Please help! :cry:
 
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Hint:
[tex]4^{x}+4^{x}=2*(4^{x})[/tex]
 
My problem isn't that simple though.
 
Is your equation
[tex]4^x + 4^x+1 = 40[/tex]
or
[tex]4^x + 4^{x+1} = 40[/tex]?
In both cases you can write [itex]4^x[/itex] as a factor.
 
Last edited:
4^x + 4^(x+1) = 40
 
Halfway done:

[tex]4^x(1 + 4^1) = 40[/tex]
 
How did you do that?
 
ms. confused said:
How did you do that?

[tex]a^{c+d} = a^c a^d[/tex]
 
OK but you only solved part of it, right?
 
  • #10
[itex]\frac{1}{16}[/itex] left to work out

[tex]4^x = 8[/tex]

[tex]4^x = 2^3[/tex]

[tex]2^{2x} = 2^3[/tex]
 
  • #11
How come it's = to 8 all in a sudden? The question says it's = to 40.
 
  • #12
Look at post #6 again. What's 1 +4?

That gives you
[tex]4^x*5=40[/tex]

Divide both sides by 5.
 

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