# Anyone good with logarithms?

1. Nov 7, 2004

### 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!

2. Nov 7, 2004

### arildno

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

3. Nov 7, 2004

### ms. confused

My problem isn't that simple though.

4. Nov 7, 2004

### 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.

Last edited: Nov 7, 2004
5. Nov 7, 2004

### ms. confused

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

6. Nov 7, 2004

### Pyrrhus

Halfway done:

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

7. Nov 7, 2004

### ms. confused

How did you do that?

8. Nov 7, 2004

### Pyrrhus

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

9. Nov 7, 2004

### ms. confused

OK but you only solved part of it, right?

10. Nov 7, 2004

### Pyrrhus

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

$$4^x = 8$$

$$4^x = 2^3$$

$$2^{2x} = 2^3$$

11. Nov 7, 2004

### ms. confused

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

12. Nov 7, 2004

### BobG

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

That gives you
$$4^x*5=40$$

Divide both sides by 5.

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