exponetial and logarithmic functions solve for x

[Louise Johnson]

Here are couple of questions I am having difficulty with. Its is just solving for x.
However neither of my answers are working out.
Thank you
Louise

Question#1

\(\displaystyle 5^{x + 3} = 8^{2x - 1}\)

My Answer:

\(\displaystyle (x + 3)\log 5 = (2x - 1)\log 8\)

\(\displaystyle x\log 5 + 3\log 5 = 2x\log 8 - \log 8\)

\(\displaystyle x\log 5 - 2x\log 8 = - \log 8 - 3\log 5\)

\(\displaystyle x(\log 5 - 2\log 8) = - \log 8 - 3\log 5\)

\(\displaystyle x = \frac{{ - \log 8 - 3\log 5}}{{\log 5 - 2\log 8}}\)

\(\displaystyle x = - 6.0982\)

Question #2

\(\displaystyle \log (x + 9) + \log x = 1\)
My answer:

\(\displaystyle 2\log x + \log 9 = 1\)

\(\displaystyle \log x + \log 9 = \frac{1}{2}\)

\(\displaystyle \log x = \frac{1}{2} - \log 9\)

\(\displaystyle \log x = .4542\)

 

[soroban]

Hello, Louise!

\(\displaystyle 1)\; 5^{x\,+\,3} \;= \;8^{2x\,-\,1}\)

My answer: \(\displaystyle \L\:x \;=\;\frac{-\log8\,-\,3\cdot\log5}{\log5\,-\,2\cdot\log8}\;\) . . . RIght!

. . . . . . . \(\displaystyle x = - 6.0982\;\) . . . no

I would simplify first . . .

\(\displaystyle \L x\;=\;\frac{\log(8)\,+\,3\cdot\log(5)}{2\cdot\log(8)\,-\,\log(5)} \;=\; \frac{\log(8)\,+\,\log(5^3)}{\log(8^2) \,-\,\log(5)} \;= \;\frac{\log(8\cdot5^3)}{\log(\frac{8^2}{3})} \;=\;\frac{\log(1000)}{\log(12.8)}\)

Therefore: \(\displaystyle \:\fbox{x\,=\,2.709513175}\)

\(\displaystyle 2)\;\;\log(x\,+\,9)\,+\,\log(x)\;= \;1\)

My answer: \(\displaystyle \:2\log(x)\,+\,\log(9)\;=\;1\;\;\) . . . NO! NO! NO!

We have: \(\displaystyle \:\underbrace{\log(x\,+\,9)\,+\,\log(x)}\;=\;1\)
. . . . . . . . . \(\displaystyle \overbrace{\log[x(x\,+\,9)]} \;=\;1\;=\;\log(10)\)

Then we have: \(\displaystyle \:x(x\,+\,9)\;=\;10\)

. . . . . . . . \(\displaystyle x^2\,+\,9x\,-\,10\;=\;0\)

. . . . . . . \(\displaystyle (x\,-\,1)(x\,+\,10)\;=\;0\)

. . which has two roots: \(\displaystyle \:x\,=\,1,\;\sout{x\,=\,-10}\)

Therefore: \(\displaystyle \,\fbox{x\:=\:1}\)

 

[Louise Johnson]

Thank you Thank you Thank you
I can't express how excited I get when I see the effort you put into a reply. It saves me days of frustration! (your work is very clear and I never have a problem understanding what you mean)I am also happy I got so far on the first one. That was alot of work for me. The second one was clearly more difficult and I am gonna literally gonna go back to studying your work.
Thank you so much
Louise