Can someone help me with Logarithms please

Show your work too ! Thanks ,

Solve for X

1) Log3 (2x + 9) = 5

2) X = ln225.9

3) 2^x + 3 = 9.2

4) log2 (2x - 3)^3 = 12

1) Log3 (2x + 9) = 5
2x+9=3^5
2x+9=243
2x=234
x=117
2) X = ln225.9
x=5.42

3) 2^x + 3 = 9.2
2^x=6.2
x=log 2(6.2)
x=2.6323

4) log2 (2x - 3)^3 = 12
3log2(2x-3)=12
log2(2x-3)=4
2x-3=2^4
2x-3=16
2x=19
x=19/2
x=9.5

3.) 2^x + 3 = 9.2 subtract 3 from both sides
2^x = 6.2 taking the log of both sides, allows you to bring the x exponent to the front
xlog2 = log6.2 divide both sides by log2 using a calc
x = log6.2/log2
x = 2.632268216

4.) log2(2x - 3)^3 = 12 you can get rid of logs by puting both sides to the base of the log number
(2x - 3)^3 = 2^12 2^12 is 4096, take the cube root of both sides
2x - 3 = 16 add 3 to both sides
2x = 19 divide both sides by 2
x = 9.5

1). 3^5= 2x+9------ 243=2x+9------234=2x------234/2=2x/2----…

2). ln is natural log and its on any scientific calculator. just type in ln(225.9) and you get your answer

3). 2^x +3= 9.2------2^x=6.2 then put in a log or ln------ ln2^x=ln6.2 place the x in front of the ln or log----xln2=ln6.2 After that IDK wat to do!

4). 2^12=(2x+3)^3 ----- 4096=(2x+3)(2x+3)(2x+3) youll need to foil the first two binomials---- 4096= 4x^2+6x+6x+9(2x+3) add like terms----4096=4x^2+12x+9(2x+3) multiply the trinomial with the binomial in parentheses----4096=8x^3+12x^2+24x+36x+1… add like terms-----4096=8x^3+12x^2+78x+27 subtract 27 from both sides of the equation----4069=8x^3+12^2+78x you cant break it down anymore

1) 2x+9 = 3^5, x = (243-9)/2
2) using calculator
3) 2^x = 9.2 - 3 = 6.2, x = ln6.2/ln2
4) 2x-3 = 2^4
2x = 19
x = 9.5

1) 2x + 9 = 3^5 = 243
2x = 234
x = 117