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

[ms. confused]

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!

 

[arildno]

Hint:
4^{x}+4^{x}=2*(4^{x})

 

[ms. confused]

My problem isn't that simple though.

 

[Galileo]

Is your equation
4^x + 4^x+1 = 40
or
4^x + 4^{x+1} = 40?
In both cases you can write 4^x as a factor.

 

[ms. confused]

4^x + 4^(x+1) = 40

 

[Pyrrhus]

Halfway done:

4^x(1 + 4^1) = 40

 

[ms. confused]

How did you do that?

 

[Pyrrhus]

How did you do that?

a^{c+d} = a^c a^d

 

[ms. confused]

OK but you only solved part of it, right?

 

[Pyrrhus]

\frac{1}{16} left to work out

4^x = 8

4^x = 2^3

2^{2x} = 2^3

 

[ms. confused]

How come it's = to 8 all in a sudden? The question says it's = to 40.

 

[BobG]

Look at post #6 again. What's 1 +4?

That gives you
4^x*5=40

Divide both sides by 5.