MHB I understanding log/exponents

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The discussion focuses on solving the equation 3(5^x) = 2^(x-2) by expressing it in the form a^x = b. The transformation involves rewriting the equation to isolate x, leading to the expression (2/5)^x = 12. Participants suggest using logarithms to simplify the equation further, which allows for isolating x effectively. The use of natural logarithms is emphasized as a common method for solving such exponent-related problems. The thread concludes with the original poster expressing gratitude and indicating a better understanding of the topic.
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I need help understand this question;

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

Any explanation/help would be amazing
 
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We are given:

$$3\cdot5^x=2^{x-2}$$

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

$$a^x=b$$

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

$$3\cdot5^x=\frac{2^x}{4}$$

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

$$12=\frac{2^x}{5^x}$$

or

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

Can you proceed?
 
micheal said:
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...
 
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!
 
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