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Year 12s: what will be the first way you research your future options? >>

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C2 exponentials and logarithms

I have to solve for x
7(4^x)=3(5^x)

Heeelp please, thanks for your time :smile:

:smile:

Original post by Senaga03

I have to solve for x
7(4^x)=3(5^x)

Heeelp please, thanks for your time :smile:

:smile:

You could take logs straight away, or alternatively:

Try and rearrange so you have the terms involving x on one side of the equation and those not involving x on the other.

7\times 4^x = 3 \times 5^x

may make it clearer as the starting point.

OP

Original post by ghostwalker

You could take logs straight away, or alternatively:

Try and rearrange so you have the terms involving x on one side of the equation and those not involving x on the other.

7\times 4^x = 3 \times 5^x

may make it clearer as the starting point.

So... I've tried, but I don't know what else to do now..

image.jpg473.8KB

Original post by Senaga03

So... I've tried, but I don't know what else to do now..

OK,

Taking logs directly - I'll give you the details:

7\times 4^x=3\times 5^x

So.

\log(7\times 4^x)=\log(3\times 5^x)

Recall that log ab = log a + log b. So,

\log 7+ \log(4^x)=\log 3 +\log(5^x)

And

\log a^b = b\log a

So, we have:

\log 7+ x\log(4)=\log 3 +x\log(5)

Now group the terms in x on one side, and non-x terms on the other. Can you take it from there?

Original post by Senaga03

I have to solve for x
7(4^x)=3(5^x)

Heeelp please, thanks for your time :smile:

:smile:

Alternative method to ghostwalker's:

Rearrange to get:

7/3 = (5/4)^x

Then take logarithms and solve for log x and hence x

(the 2 methods should both lead to the same answer)

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