# Thread: Exponential Equations with Logarithms

1. ## Exponential Equations with Logarithms

Hi everyone,
I need some help with these exponential equations. I am required to use logarithms to solve them. I have hit a wall!

1. 2^x=30
2. 5^x-1=3^x
3. 3.5^3x+1=65.4
4. 16^x-4=3^3-x
5. 7^x-2=5^x

Help/explanations would be greatly appreciated!
Thanks so much!

2. Please learn LaTeX, or at least use grouping symbols. For instance, in #2 is it supposed to be
$\displaystyle 5^x - 1 = 3^x$ or $\displaystyle 5^{x-1} = 3^x$?

I'll do the first, as it doesn't look ambiguous. To isolate the x, take the log base-2 of both sides:
\displaystyle \begin{aligned} 2^x &= 30 \\ \log_2 2^x &= \log_2 30 \\ x &= \log_2 30 \end{aligned}

If you need to state your answer as a decimal, use the change of base formula to change into either natural logs (base e) or common logs (base 10), and then plug into the calculator:
$\displaystyle \log_2 30 = \dfrac{\log 30}{\log 2} \approx 4.907$
OR
$\displaystyle \log_2 30 = \dfrac{\ln 30}{\ln 2} \approx 4.907$

3. Originally Posted by allegroconbrio95
Hi everyone,
I need some help with these exponential equations. I am required to use logarithms to solve them. I have hit a wall!

1. 2^x=30
2. 5^x-1=3^x
3. 3.5^3x+1=65.4
4. 16^x-4=3^3-x
5. 7^x-2=5^x

Help/explanations would be greatly appreciated!
Thanks so much!
Hi allegroconbrio95,

[1] Using natural log:

$\displaystyle 2^x=30$

$\displaystyle \ln 2^x= \ln 30$

$\displaystyle x\ln 2=\ln 30$

$\displaystyle x=\frac{\ln 30}{\ln 2}$

$\displaystyle x \approx 4.9069$

[2] $\displaystyle 5^{x-1}=3^x$

$\displaystyle \ln 5^{x-1}=\ln 3^x$

$\displaystyle (x-1)\ln 5 = x \ln 3$

$\displaystyle x \ln 5-\ln 5=x \ln 3$

$\displaystyle x \ln 5 - x \ln 3 = \ln 5$

$\displaystyle x(\ln 5 - \ln 3)= \ln 5$

$\displaystyle x=\frac{\ln 5}{\ln 5-\ln 3}$

$\displaystyle x \approx 3.1507$

Now, the rest are similar to [2] so you have a pretty good model to use. Good luck.