Log equation help

**200001** · December 20th 2009, 04:56 AM

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

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

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

My math type didnt import the text in

Thank you

**skeeter** · December 20th 2009, 05:07 AM

Originally Posted by 200001

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

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

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

My math type didnt import the text in

Thank you

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

$\log\left(7^{x-3}\right) = \log\left(4^{2x}\right)<br />$

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

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

$x\log{7} - 2x\log{4} = 3\log{7}<br />$

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

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

**200001** · December 20th 2009, 05:49 AM

Thank you very much, I see where I went wrong

What do I do with this one?

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

Thanks

**HallsofIvy** · December 20th 2009, 06:06 AM

Originally Posted by 200001

Thank you very much, I see where I went wrong

What do I do with this one?

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

Thanks

Do you mean $2^x2^{x+1}= 10$ ?
(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": $2^x2^{x+1}= 2^{x+ x+ 1}= 2^{2x+1}$ and then take the logarithm of both sides.

**200001** · December 21st 2009, 06:41 AM

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?

**skeeter** · December 21st 2009, 07:16 AM

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

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

now you have the equation ...

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

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

$\pm 3 = \log_5{x}$

$x = 5^3$ or $x=5^{-3}$

**200001** · December 21st 2009, 09:15 PM

Much obliged, thanks

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