1. Logs

How to you answer the following question using logs:

What are the two solutions for x of 2^(2x)-8*2^x+15=0
In words: 2 to the power 2x, -8 multiplied by 2 to the power x, plus 15.

2. Originally Posted by haku
How to you answer the following question using logs:

What are the two solutions for x of 2^(2x)-8*2^x+15=0
In words: 2 to the power 2x, -8 multiplied by 2 to the power x, plus 15.

$\displaystyle 2^{2x} - 8*2^x + 15 =(2^x)^2 - 8*2^x + 15=0$
$\displaystyle let \, u=2^x$

then this is just a quadratic equation in u, that is

$\displaystyle u^2 - 8u +15 = (u-3)(u-5)=0$

so u=3 or u=5
but since

$\displaystyle u= 2^x$

therefore $\displaystyle x=\frac{\ln 3}{\ln 2} = \log_2 3 \, or \, x=\frac{\ln 5}{\ln 2} = \log_2 5$

3. Thanks for the help and your explanation

4. From $\displaystyle 2^x=3\implies x=\log_23$ & $\displaystyle 2^x=5\implies x=\log_25.$

From this way, it's easier to check the solutions without a calculator.

--

$\displaystyle 2^{2x}-8\cdot2^x+15=0.$

So for $\displaystyle x=\log_23,$

$\displaystyle 2^{2x}=2^{\log_29}=9$ & $\displaystyle 2^x=3,$ therefore

$\displaystyle 9-8\cdot3+15=0.$

You can do the same procedure to verify the second solution.

5. What about questions in the form:
log(base6)4 + log(base6)9 + log(base3)45 -log(base3)5

Thanks

6. Originally Posted by haku
What about questions in the form:
log(base6)4 + log(base6)9 + log(base3)45 -log(base3)5

Thanks
It would be log(base 6) 36 + log(base 3) 9