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

  • Thread starter ms. confused
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    Logarithms
In summary, The conversation is about solving the exponential equation 4^x + 4^x+1 = 40 using logarithms. The expert suggests rewriting 4^x as a factor and then dividing both sides by 5, leaving the equation 4^x = 8. They then explain how 8 can be written as 2^3 and how to further simplify the equation to 2^{2x} = 2^3. The person seeking help is confused about how the equation suddenly becomes equal to 8 when the question states it is equal to 40. The expert clarifies that 1+4=5 and the equation becomes 4^x*5 = 40.
  • #1
ms. confused
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0
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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  • #2
Hint:
[tex]4^{x}+4^{x}=2*(4^{x})[/tex]
 
  • #3
My problem isn't that simple though.
 
  • #4
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:
  • #5
4^x + 4^(x+1) = 40
 
  • #6
Halfway done:

[tex] 4^x(1 + 4^1) = 40 [/tex]
 
  • #7
How did you do that?
 
  • #8
ms. confused said:
How did you do that?

[tex] a^{c+d} = a^c a^d [/tex]
 
  • #9
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.
 

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

1. What are logarithms and how do they work?

Logarithms are mathematical functions that help us solve equations with exponents. They work by taking a number as input and returning the exponent needed to produce that number. For example, if the base of the logarithm is 10, log(100) = 2 because 10^2 = 100.

2. Why are logarithms useful?

Logarithms are useful because they allow us to convert complex calculations involving exponents into simpler addition and subtraction problems. They also help us compare numbers that vary over a large range, such as in scientific measurements.

3. How do I solve equations with logarithms?

To solve equations with logarithms, you can use the exponent property of logarithms, which states that log(a^b) = b*log(a). This allows you to move the exponent of the unknown variable to the other side of the equation and solve for the variable.

4. Can you give an example of a real-life application of logarithms?

Logarithms are used in many fields, such as engineering, economics, and astronomy. One example of a real-life application is the Richter scale, which measures the intensity of earthquakes using logarithms. Each whole number increase on the scale represents a 10-fold increase in earthquake intensity.

5. Are there any common mistakes to avoid when working with logarithms?

One common mistake when working with logarithms is forgetting to include the base of the logarithm in the final answer. Another mistake is mixing up the base of the logarithm when solving equations, which can result in incorrect solutions. It is important to double check your work and make sure you are using the correct base for the given problem.

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