SOLUTION: 1. 5+2logx=4 2. log(x-2)+log(2x-3)=2logx 3. 4^x+3=7^x 4. log2x + log2 (x+2)=log2 (x+6) 5. 5^(x/2)=123

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 1. 5+2logx=4 2. log(x-2)+log(2x-3)=2logx 3. 4^x+3=7^x 4. log2x + log2 (x+2)=log2 (x+6) 5. 5^(x/2)=123      Log On


   



Question 393281: 1. 5+2logx=4
2. log(x-2)+log(2x-3)=2logx
3. 4^x+3=7^x
4. log2x + log2 (x+2)=log2 (x+6)
5. 5^(x/2)=123

Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Add -5 to both sides:



Multiply both sides by



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John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. 5+2logx=4
2logx = -1
log(x) = -1/2
x = 10^-1/2
x = (1/10)sqrt(10)
==============================
2. log(x-2)+log(2x-3)=2logx
log[(x-2)(2x-3)] = log(x^2)
---
2x^2-7x+6 = x^2
x^2-7x+6 = 0
(x-6)(x-1) = 0
x = 6 is a usable answer; x = 1 is not.
==========
3. 4^(x+3)=7^x
(x+3)log(4) = xlog(7)
x[log(7)-log(4)] = 3log(4)
---
x = [3log(4)]/[log(7)-log(4)]
---
x = 7.4317
============
4. log2x + log2 (x+2)=log2 (x+6)
---
log2[x(x+2)] = log2(x+6)
----
x(x+2) = x+6
x^2+x-6= 0
(x+3)(x-2) = 0
-----
Positive solution:
x = 2
=======================
5. 5^(x/2)=123
(x+2)log(5) = log(123)
x+2 = log(123)/log(5)
---
x+2 = 2.99
x = 0.99
===============
Cheers,
Stan H.