Log equation help

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• Dec 20th 2009, 04:56 AM
200001
Log equation help
Please can someone take me through the steps in this as I have the x's on both sides

$\displaystyle 7^(x-3)=4^(2x)$

The (x-3) and (2x) are exponents

My math type didnt import the text in (Headbang)

Thank you
• Dec 20th 2009, 05:07 AM
skeeter
Quote:

Originally Posted by 200001
Please can someone take me through the steps in this as I have the x's on both sides

$\displaystyle 7^(x-3)=4^(2x)$

The (x-3) and (2x) are exponents

My math type didnt import the text in (Headbang)

Thank you

$\displaystyle 7^{x-3} = 4^{2x}$

$\displaystyle \log\left(7^{x-3}\right) = \log\left(4^{2x}\right)$

$\displaystyle (x-3)\log{7} = 2x\log{4}$

$\displaystyle x\log{7} - 3\log{7} = 2x\log{4}$

$\displaystyle x\log{7} - 2x\log{4} = 3\log{7}$

$\displaystyle x(\log{7} - 2\log{4}) = 3\log{7}$

$\displaystyle x = \frac{3\log{7}}{\log{7} - 2\log{4}}$
• Dec 20th 2009, 05:49 AM
200001
Thank you very much, I see where I went wrong

What do I do with this one?

$\displaystyle 2^x multiplied by 2^(x+1) = 10$

Thanks
• Dec 20th 2009, 06:06 AM
HallsofIvy
Quote:

Originally Posted by 200001
Thank you very much, I see where I went wrong

What do I do with this one?

$\displaystyle 2^x multiplied by 2^(x+1) = 10$

Thanks

Do you mean $\displaystyle 2^x2^{x+1}= 10$?
(The code for that is "2^x2^{x+1}= 10". Notice the "{ }" to group x+ 1 as a single exponent.)

Use a "law of exponents": $\displaystyle 2^x2^{x+1}= 2^{x+ x+ 1}= 2^{2x+1}$ and then take the logarithm of both sides.
• Dec 21st 2009, 06:41 AM
200001
Great, yes I get that

Now I am on these ones

9log (base x of 5) = log (base 5 of x)

not sure how i get through these as Im trying to change the base I believe?
• Dec 21st 2009, 07:16 AM
skeeter
Quote:

Originally Posted by 200001
Great, yes I get that

Now I am on these ones

9log (base x of 5) = log (base 5 of x)

not sure how i get through these as Im trying to change the base I believe?

in future, start a new problem with a new thread.

using change of base, note that

$\displaystyle \log_x(5^9) = \frac{\log_5(5^9)}{\log_5{x}} = \frac{9}{\log_5{x}}$

now you have the equation ...

$\displaystyle \frac{9}{\log_5{x}} = \log_5{x}$

$\displaystyle 9 = (\log_5{x})^2$

$\displaystyle \pm 3 = \log_5{x}$

$\displaystyle x = 5^3$ or $\displaystyle x=5^{-3}$
• Dec 21st 2009, 09:15 PM
200001
Much obliged, thanks (Nod)