SOLUTION: Having difficulty with two problems:
2^(x+3)=5^x
and
10e^(3x-7)=5
For 2^(x+3)=5^x I converted it into a log with base of 2
log(base2) (2^(x+3)) = log(base2) 5^x
x
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Having difficulty with two problems:
2^(x+3)=5^x
and
10e^(3x-7)=5
For 2^(x+3)=5^x I converted it into a log with base of 2
log(base2) (2^(x+3)) = log(base2) 5^x
x
Log On
Question 65026: Having difficulty with two problems:
2^(x+3)=5^x
and
10e^(3x-7)=5
For 2^(x+3)=5^x I converted it into a log with base of 2
log(base2) (2^(x+3)) = log(base2) 5^x
x+3=log(base2) 5^x
x=log(base2) (5^x) - 3
and get stuck there.. am I even on the right track? As for the 2nd one, I've made no progress at all.
You can put this solution on YOUR website! Having difficulty with two problems:
2^(x+3)=5^x
Take the log of both sides to get:
(x+3)log2 = xlog5
Distribute to get:
(log2)x + 3log2 = xlog5
Get the x-terms together as follows:
(log2-log5)x=-3log2
Evaluate to get:
-0.39794..x=-0.903089987..
Divide to get:
x=2.269...
-----------------
and
10e^(3x-7)=5
Divide by 10 to get:
e^(3x-7)=(1/2)
Take the natural log of both sides to get:
3x-7=ln(1/2)
3x=-0.693147...+7
3x=6.30685...
x=2.10228...
Cheers,
Stan H.
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