1. marco26

just use $a ^{x}=y$ ->$\log _{a}y=x$

2. Owlcoffee

$(3)(5^x)= 48$ When we want to solve an exponential equation, called like this because the variable right now is in the exponent, we have to use the reciprocate operation in order to bring the x down as a normal variable and not an exponent. What is reciprocate to exponents?, well Logarithms. In essence, we can use any logarithm we want, but we always take the logarithm with the convenient base in order to make things simple, so, we will can begin by getting rid of the 3 by dividing both sides by 3: $5^x=\frac{ 48 }{ 3 }$ $5^x=16$ So, now, what we will do is take the logarithm with base 5, which means: $$\log_{5}$$ on both sides of the equation, further on, use the property of logarithms $$\log_{a} b^n=n \log_{a} b$$ to bring the variable "x", down from the exponent: $5^x=16$ $\log_{5} 5^x=\log_{5} 16$ $$x \log_{5} 5=\log_{5} 16$$ ...and $$\log_{5} 5=1$$ so therefore: $x=\log_{5} 16$

3. anonymous

Thank you so much for taking the time out of your day to help me it it means a lot.

4. anonymous

@Owlcoffee

5. Owlcoffee

No problem

6. Jhannybean

$3(5^x) = 48$Dividing both sides by 5, we would get $5^x = \frac{48}{3} \qquad \implies 5^x = 16$ and now we can take the natural log of both sides to find x. $\log_5 (5^x) = \log_5 (16)$$x = \log_5(16)$ Using the change of base formula we can simplify $$\log_5(16)$$ into a fractional exponent. Remember that $$\log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}$$ Applying this you would get $\color{red}{\boxed{x=\log_5(16) = \frac{\log(2^4)}{\log(5)}=\frac{4\log(2)}{\log(5)}}}$

7. Jhannybean

dividing both sides by 3*