# Homework Help: Simple Indices

1. May 4, 2010

### Champdx

1. The problem statement, all variables and given/known data
2^x+3^x=13, x=2. How do i prove it?

3. The attempt at a solution
i did this
x log 2 + x log 3 = log 13
x(log 2 + log 3)=log 13
x(0.301+0.477)= 1.11
0.778x=1.11
x=1.43

2. May 4, 2010

### HallsofIvy

??? Why not just substitute "2" for "x"?

$2^x+ 3^x= 2^2+ 3^2= 4+ 9= 13$.

You are not asked so solve the equation!

3. May 4, 2010

### tiny-tim

Hi Champdx!

(try using the X2 tag just above the Reply box )
Nooo

that's log (x2x3)

4. May 4, 2010

### Champdx

Actually i know that the answer x=2 but how am i suppose to prove it?

5. May 4, 2010

### tiny-tim

If the question says "solve 2x + 3x = 13", then there's no exact way of doing it, you'll have to use an approximation method (or guess).

If the question says "show that 2x + 3x = 13 has a solution x = 2", or "prove that 2x + 3x = 13 has a solution x = 2", then all you need to do is to show that 22 + 32 = 13.

6. May 4, 2010

### Staff: Mentor

You can't get the equation above from the one you started with. log(A + B) $\neq$ logA + logB. What you did was to take the log of both sides to get
log(2^x + 3^x) = log 13. That's a legitimate step, but it doesn't lead you anywhere.
The problem is that log(2^x + 3^x) $\neq$ log 2^x + log 3^x.

7. May 5, 2010

### Champdx

Is there any way to solve this question by calculation?

Thanks.

8. May 5, 2010

Nope.

9. May 5, 2010

### The Chaz

Champ, read this quote again. There are only two options:
1) "Solve..." or
2) "Show that"/"Prove..."

There's a phrase that comes up frequently in math (that honestly kinda p*sses me off!)...
"By inspection...".
This means "I can SEE the answer, but I can't/won't show HOW you could get the same answer without guessing". There might be a better description out there for this phrase, which I think applies to our situation.

We can see that x = 2 is a solution. We can see that it is a UNIQUE solution because increasing/decreasing x will increase/decrease BOTH terms on the left hand side of the equation...