SOLUTION: Solve the exponential equation algebraically.
5^x+2=4^1-x
Here is what I tried:
log 5 ^x+2= log 4^1-x
Then the exponent goes in front of the log
(x+2) log 5=(1-x) log 4
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Solve the exponential equation algebraically.
5^x+2=4^1-x
Here is what I tried:
log 5 ^x+2= log 4^1-x
Then the exponent goes in front of the log
(x+2) log 5=(1-x) log 4
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Question 1059931: Solve the exponential equation algebraically.
5^x+2=4^1-x
Here is what I tried:
log 5 ^x+2= log 4^1-x
Then the exponent goes in front of the log
(x+2) log 5=(1-x) log 4
I get confused after this step. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Solve the exponential equation algebraically.
5^x+2=4^1-x
Here is what I tried:(log 5 ^x+2= log 4^1-x
Then the exponent goes in front of the log
(x+2) log 5=(1-x) log 4
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(x+2)/(1-x) = log(4)/log(5) = 0.8614
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(x+2) = 0.8614- 0.8614x
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1.8614x = -1.1386
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x = -0.6117
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Cheers,
Stan H.
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