MHB Generating a linear equation from the given expression

[nicodemus1]

Good Day,

I need some help with the following problem.

I do not know how to generate a linear equation connecting y and x from the expression below as the 4 seems to complicate things

\(3^y = 4(3^{x-2}) - 1\)

Could some one please care to give me some advice?

Thanks in advance.

 

[Evgeny.Makarov]

What complicates things is -1. Without it, taking log₃ of both sides we get y = log₃4 + x - 2. As it is, the relationship between x and y is not linear.

 

[SuperSonic4]

Good Day,

I need some help with the following problem.

I do not know how to generate a linear equation connecting y and x from the expression below as the 4 seems to complicate things

\(3^y = 4(3^{x-2}) - 1\)

Could some one please care to give me some advice?

Thanks in advance.

By linear equation I take you mean something which will give a straight line when plotted on a graph? Even so that $-1$ is much more problematic than the 4 because it means we can't use logarithms effectively (for the record the 4 wouldn't be a problem thanks to the log addition law) so a log-log graph won't come out linear either.

 

[nicodemus1]

Thank you all for your advice.

However, I'm not required to transform the expression into a linear equation to plot it. I have to solve for x and y.

I apologise for not making myself clearer earlier. I was given 2 equations and I have to solve for x and y.

The other equation is: \(64(4^y) = 16^x\).

I had no problems in obtaining an equation connecting x and y from the above equation. It is the 2nd equation that I mentioned earlier in this post that I'm having trouble with in getting an equation connecting x and y.

Hope this information helps you give me a clearer direction.

 

[SuperSonic4]

Thank you all for your advice.

However, I'm not required to transform the expression into a linear equation to plot it. I have to solve for x and y.

I apologise for not making myself clearer earlier. I was given 2 equations and I have to solve for x and y.

The other equation is: \(64(4^y) = 16^x\).

I had no problems in obtaining an equation connecting x and y from the above equation. It is the 2nd equation that I mentioned earlier in this post that I'm having trouble with in getting an equation connecting x and y.

Hope this information helps you give me a clearer direction.

Ah, so it's simultaneous equations? This makes life much easier. The equation in the above post is also easier to manipulate.

$64 \cdot 4^y = 16^x$

Note that 64, 4 and 16 are all powers of 4 so the equation can be written as

$4^3 \cdot 4^y = (4^2)^x$

From the exponent laws: $4^{3+y} = 4^{2x}$

We also know that if the bases are the same the exponents must also be the same (or take logs to base 4, your choice) leading to: $3+y = 2x$

It is trivial to make y either y or x the subject of that equation and then to substitute it into $3^y = 4(3^{x-2})-1$

 

[chisigma]

Thank you all for your advice.

However, I'm not required to transform the expression into a linear equation to plot it. I have to solve for x and y.

I apologise for not making myself clearer earlier. I was given 2 equations and I have to solve for x and y.

The other equation is: \(64(4^y) = 16^x\).

I had no problems in obtaining an equation connecting x and y from the above equation. It is the 2nd equation that I mentioned earlier in this post that I'm having trouble with in getting an equation connecting x and y.

Hope this information helps you give me a clearer direction.

If Your first equation is...

$\displaystyle 3^{y}= 4\ (3^{x-2})-1$ (1)

... with simple steps it becomes...

$\displaystyle y=\frac{\ln (\frac{4}{9}\ 3^{x}-1)}{\ln 3}\ $ (2)

Now You insert (2) into $\displaystyle 64\ (4^{y})= 16^{x}$ and solve in x...

Kind regards

$\chi$ $\sigma$