Check the following logarithmic equation I solved please!

Question

log(x+1)+log(e^x)=10

(x+1)e^x=10¹⁰

x=W₀(10¹⁰e)-1

x=19.98

ln(x+1)+log(x+1)+log₄(x+1)=10

x+1=10^x=10^{{20log(e)log(2)}/{2log(e)log(2)+log(4e)}}

^x=10^{{20log(e)log(2)}/{2log(e)log(2)+log(4e)}-1}

^x=102.44

Let me know if this is correct or if there is a way to solve the first without the Lambert W function. The initial question is bolded and in italics.

Answer 1

Log(x+1) + log(e^x) = 10

logs adding = insides multiply

log[ (x+1)(e^x) ] = 10

raise everything with base 10 to cancel out the log

(x+1)(e^x) = 10^10

This one can’t be solved manually in any other way so I solved graphically by graphing y = (x+1)e^x and y = 10^10 and finding their point of intersection

x = 19.983

Ln(x+1) + log(x+1) + log_4(x+1) = 10

apply change of base so all logs are base 10

Log(x+1)/log(e) + log(x+1) + log(x+1)/log(4) = 10

(remember log e and log 4 are just numbers)

I am going to move 1/log(e) and 1/log(4) into the logs as the exponent of the inside x+1

log[ (x+1)^(1/log(e)) ] + log(x+1) + log [ (x+1)^(1/log(4)) ] = 10

now the logs have no coefficients and are all adding with the same base = combine

*the exponents will all add together*

LOG[ (x+1)^(1/log(e) + 1 + 1/log(4)) ] = 10

raise everything to the base 10 to cancel the log

(x+1)^(1/log(e) + 1 + 1/log(4)) = 10^10

Root everything by 1/log(e) + 1 + 1/log(4) to get the x + 1 alone (it’s roughly rooting by 4.963549 and it is going to divide the power 10 on the right side)

x + 1 = 10^(10/4.963549)

x = 10^2.041687 - 1

x = 102.440