please help asap LOGARITHMS EQUATION

[mathsem]

given that log a = 5
solve:
a^2x -14a^x + 40 = 0

I keep getting the answer of 0.03 but there’s meant to be two answers and idk how!
The answers in my textbook are x=0.2 or 0.12
how???

 

[tkhunny]

If only you has shown us how you managed that...

How would you solve this: x^2 - 14x + 40 = 0 ?

 

[pka]

given that log a = 5
solve: a^2x -14a^x + 40 = 0
I keep getting the answer of 0.03 but there’s meant to be two answers and idk how!
The answers in my textbook are x=0.2 or 0.12 how???

It should be noted that base ten, \(\displaystyle \log_{10}(a)=5\).
\(\displaystyle a^{2x}-14a^x+40=(a^x-10)(a^x-4)=0\)

 

[mathsem]

Thank you! I have reached the answers now ?

 

[ronaledsmith]

Thanks for this help, I'm also searching for that.