SOLUTION: Hi. Desperately need help with these. I think I got the answer to one, but am totally stuck on the other. Here they are: Solve 4^x+2 = 9^x-1 Solve 7^x+2 = 10^x-3 (I got 3

Question 81965: Hi. Desperately need help with these. I think I got the answer to one, but am totally stuck on the other. Here they are:
Solve 4^x+2 = 9^x-1
Solve 7^x+2 = 10^x-3 (I got 3+2log7 ((over)) 1-log7) Is that right?
Please help!! Thanks so much!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Solve 4^x+2 = 9^x-1
:
Assume you mean:

:
Write the log equiv of exponents:
(x+2)*log(4) = (x-1)* log(9)
:
Find the log of 4 and 9
.602(x+2) = .954(x-1)
:
Multiply what's inside the brackets and solve for x:
.602x + 1.204 = .954x - .954
:
1.204 + .954 = .954x - .602x
:
2.158 = .352x
:
x = 2.158/.352
:
x = 6.13
:
:
Check solution on a calc:
4^8.13 = 78478
and
9^5.13 = 78571, not exact because of rounding off
:
:
:
Solve 7^x+2 = 10^x-3
:
Do this one exactly the same way:

:
.845(x+2) = 1(x-3); log of 10 is 1
:
.845x + 1.69 = x - 3
:
1.69 + 3 = x - .845x
:
4.69 = .155x
:
x = 4.69/.155
:
x = 30.258
:
Check solution on a calc
7^32.258 = 1.82(10^27)
10^27.258 = 1.81(10^27)
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