I understanding log/exponents

[micheal1]

I need help understand this question;

3(5^x)=2^(x-2)

Any explanation/help would be amazing

 

[MarkFL]

We are given:

\(\displaystyle 3\cdot5^x=2^{x-2}\)

Our goal here is to express this equation in the form:

\(\displaystyle a^x=b\)

So, in light of this, let's write the equation as:

\(\displaystyle 3\cdot5^x=\frac{2^x}{4}\)

Now, if we multiply both sides of the equation by \(\displaystyle \frac{4}{5^x}\) we obtain:

\(\displaystyle 12=\frac{2^x}{5^x}\)

or

\(\displaystyle \left(\frac{2}{5}\right)^x=12\)

Can you proceed?

 

[Prove It]

I need help understand this question;

3(5^x)=2^(x-2)

Any explanation/help would be amazing

As elegant as Mark's solution is, it is not entirely obvious that that is what could be done.

A good strategy with these sort of problems, where you are trying to solve for an exponent, is to take the logarithm of both sides (any base is fine, but the natural logarithm is the most commonly used), so

$\displaystyle \begin{align*} 3 \cdot 5^x &= 2^{x- 2} \\ \ln{ \left( 3 \cdot 5^x \right) } &= \ln{ \left( 2^{x - 2} \right) } \\ \ln{(3)} + \ln{ \left( 5^x \right) } &= \left( x-2 \right) \ln{ (2) } \\ \ln{(3)} + x\ln{ (5)} &= x\ln{(2)} - 2\ln{(2)} \\ x\ln{(2)} - x\ln{(5)} &= \ln{(3)} + 2\ln{(2)} \\ x \left[ \ln{(2)} - \ln{(5)} \right] &= \ln{(3)} + 2\ln{(2)} \end{align*}$

Finish it...

 

[micheal1]

Thanks heaps guys!, I think I understand it now~ I'll create new one's in a similar format a revise a little now ahaha.

Thanks!